What two digit number is three times the sum of its digits?
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153 = 1 3 + 5 3 + 3 3 . Also:. 370 = 3 3 + 7 3 + 0 3 . 371 = 3 3 + 7 3 + 1 3 . 407 = 4 3 + 0 3 + 7 3 . See Related Links . See the Related Links for "Exponentiated digit sums" to the bottom for more...
The number is 45. The sum of its digits i.e. 4+5=9 Five times the sum of its digits i.e. 5 times 9 which is 45
that doesn't make a whole lot of sense, cuz the only digit of six is 6, so, if I'm interpreting this question correctly, the answer would just be 36. no, i actually think its 54.
Nine hundred thirty (930) is the highest three digit number whose digits have a sum of 12. 9+3+0=12. Any numbers higher than that will give you an sum greater than 12.
The sum of the digits of a two-digit number is 14 The first digit is 4 less than twice the second digit What is he number?
86 let number=ab; sum of digits=a+b=14 -- eq(1) and give first digit is 4 less than twice of 2nd.so a+4=2b --eq(2) solve 1 and 2 u can get, a=8 and b=6 so number is 86 Also, using Chagorian's Degharvium, you can find out how ab 2 is equal to N 1 plus Y 4 divided by a 3 plus b 4 . If you know wh…at Chagorian's Degharvium is, you can do it fast and easily. (MORE)
What is the number of a two digit number that is four times the sum of its digits The tens digit is 3 less than the units digit?
36.  b â 3 = a  4(a + b) = 10a + b  4a + 4b = 10a + b Substitute  (b â 3) for a in :  4(b â 3) + 4b = 10(b â 3) + b  4b â 12 +4b = 10b â 30 + b  18 = 3b  6 = b  3 = a
There exist no numbers which are less than the sum of their digits. This is obvious for single digit numbers, since each number would be exactly equal to the sum of its digits. For multi-digit numbers, this still holds true. Let's take a two digit number in the form: ab. Where a and b represen…t the digits of this number. The sum of the digits of ab = a + b. The number ab = a*10 + b. It should now be obvious that ab will always be greater than a + b. For larger numbers, the difference between the sum of digits and the number only grows, as a number abc = a*100 + b*10 + c, abcd = a*1000 + b*100 + c*10 + d, etc. Proof by exhaustion (via Java code, since this is listed in the programming category): for (int n = 10; n = sumDigits for all 10 (MORE)
If there are two digits than the first one is ten times its value. So as an algebraic equation we get 10A + B - 3 = 7(A x B) This can be simplified to 10A + B - 3 = 7A + 7B and this reduces to 10A - 7A = 7B - B +3 or 3A =6B +3 A solution to this is 3(3) = 6(2) +3 Which …would make the original equation 32 = 7(3+2)-3 (MORE)
What is the answer four times the sum of the digits of a two-digit numbr is equal to the numberif the digits are reversed th resulting number is 27 greater than the original number what is the number?
Name the digits 'T' and 'U', for 'tens digit' and 'units digit'.. First statement: 4 (T + U) = 10T + U. 4T + 4U = 10T + U > 3U = 6T > U = 2T . Second statement: 10U + T = 10T + U + 27. 9U = 9T + 27 > U = T + 3 . === . U = 2T and U = T + 3 > 2T = T + 3 > T = 3 . U = 2T = 6 . The number could… be 36 .. ===. Check:. First statement: 4 (T + U) = 10T + U. 4 (3 + 6) ?=? 10(3) + 6. 4(9) ?=? 30 + 6. 36 = 36 Yay !. Second statement: 10U + T = 10T + U + 27. 10(6) + 3 ?=? 10(3) + 6 + 27. 60 + 3 ?=? 30 + 6 + 27. 63 = 63 Double yay !. The number is definitely 36 . (MORE)
That can be solved by treating the digits of the number as two separate variables, and then expressing the question as an equation: 10A + B = 3AB 3AB - 10A - B = 0 A(3B - 10) - B = 0 A = B/(3B - 10) Now, we know that A and B must both be positive single digit numbers (otherwise, our firs…t equation wouldn't fit the problem). That means "3B - 10" must give us a positive (in order to get that positive result). Because of that, we can see that B must be greater than or equal to 4. So let's try plugging in some values: if B = 4: A = 4/(12 - 10) = 2 if B = 5: A = 5/(15 - 10) = 4/5 As B goes up, A will get smaller and smaller. Since it is already a fraction, we know that the first answer is the only one. The answer then is 24. (MORE)
There are 7 possible answers: 39, 48, 57, 66, 75, 84, 93. Answer can't include a 0,1 or 2, since the other digit will need to have a value greater than 9.
The sum of the digits in two- digit number is 15 If the number is 6 more than 15 times the units digit find the number?
If the sum of the digits in a two-digit number is 15, and the number is 6 more than 15 times the units digit, the number is 96.. Let A and B be the digits. (A is the tens digit and B is the units digit) . A + B = 15, therefore A = 15 - B. 10A + B = 15B + 6 Substituting for A, and solving for B…... 10(15 - B) + B = 15B + 6 150 - 10B + B = 15B + 6 150 - 9B = 15B + 6 144 - 9B = 15B 144 = 24B B = 6. Back substitute B into first equation and solve for A... A + 6 = 15 A = 9. Therefore, the digits are 9 and 6, so the number is 96. (MORE)
What is the number if there is a 4 digits number and no number is repeated The digit in the tens place is three times the digit in the thousand place. The sum of the digits in the number is 27?
3897 7 is in ones place 9 is in tens place (and is three times the number in thousands place) 8 is in the hundreds place 3 is in the thousands place 7+9+8+3=27
In a certain two-digit number the tens' digit exceeds the units' by 5 What is the number if it is known that the number is equal to ten times the sum of the digits?
The two digit numbers in which the tens digit is five greater than the units digit are: . 50 . 61 . 72 . 83 . 94 . The number we are looking for is ten times the sum of its digits, and so must be a multiple of 10. The only one of the above numbers which fits this is 50 , as 5 is five grea…ter than 0, and 50 = 10 x (5+0). (MORE)
Multiply the three-digit number by the one's digit, or last digit, of the two-digit number. That is your first part. Now multiply by the second-to-last digit, or ten's digit, and multiply the result by 10. That is your second part. Add the two parts and that is your answer.
What is a number between 75 and 100 whose tens digit is three times the ones digit and the sum of the digits is five and the hundreds digit and the ones digit are the same?
There is no number. 100 is the only one that has a hundreds digit. It's impossible.
You are a three digit odd number your tens digit is one less than your hundredths digit but two more than your ones digit the sum of your digits is twenty who are you?
x = 100A + 10B + C A = B + 1 B = C + 2 â´A = C + 3 A + B + C = 20 â´(C + 3) + (C + 2) + C = 20 3C = 15 C = 5 â´ B = 7 â´ A = 8 The number is 875
The sum of all the integers between 100 and 999 is equal to 494450 - which also happens to be the answer to the sum 899 x 500 (which refers to the number of numbers multiplied by the mean number + 0.5).
The highest 3-digit number is 999. It's the 999 th counting number. The first 99 counting numbers have fewer than 3 digits. So there are (999 - 99) = 900 positive 3-digit numbers. The first is 100, the last is 999. 100 + 999 = 1,099 101 + 998 = 1,099 102 + 997 = 1,099 . . . There are (900/2) =… 450 pairs; each pair adds up to 1,099. The sum of all pairs is (450 x 1,099) = 494,550 . (MORE)
The ten's digit of a two-digit number is 3 more than the units' digit The number is 8 more than six times the sum of the digits Find the number?
Let x be the unknown number and let y be the ten's digit, then x = 10y + y - 3 = 11y - 3 , and x = 8 + 6(y + y - 3) = 8 + 6(2y - 3) = 8 + 12y - 18 = 12y -10 , therefore 12y - 10 = 11y - 3. Subtracting 11y and adding 10 to each side, y = 7. Since the ten's digit is 7, the unit's digit must b…e 4, and the number is 74. (MORE)
The only two digit numbers whose digits add up to 17 are 89 and 98. Since 98 is an even number, 89 is prime.\n
You see how many times the two digit number goes into the first two numbers then you will take them away and then you will get the third number and put it on the end of your answer then you will see how many times your two digit number goes into your new number and then you will get your answer. And… by the way I am eleven and I am still in primary school so you should really know it. I can divide a two digit by a seventeen digit number and that ins easy peasy. (MORE)
The tenths digit of a two-digit number is 3 more than the units' digit. The number is 8 more than six times the sum of the digits. Find the number.?
74 Let a 1 represent the tens digit and a o represent the unit digit. From the problem, we have two equations: 10a 1 +a o = 8+6(a 1 +a o ) a 1 =3+a o Substitute a 1 from the second equation into the first equation and solve for a 0 : 10(3+a o )+a o = 8+6((3+a o )+a o ) 30+10a o +a o =… 8+18+6a o +6a o 30+11a o = 26+12a o a 0 =4 Substituting this back into the second equation, we get: a 1 =3+4=7 (MORE)
There are actually three valid answers: 17, 53, and 71 All are prime and all have digits adding to 8.
Call the digits 'A' and 'B', so that the number is written: AB. 10A + B = 4 (A+B) 10A + B = 4A + 4B 6A = 3B 2A = B ---> any 2-digit number where the units digit is twice the tens digit 12 or 24 or 36 or 48
Half of 382 is 181. 181 can't be evenly divided by any number except 1 and 181. (It's a prime number.) So 362 can't have any 2-digit factor. The question has no solution.
The sum of the digits of a two digit number is 12 the tens digit is twice the ones digit what is the two digit number?
The number is 84 . Assign the digits in the number the letters a and b. GIVEN: a + b = 12 GIVEN: a = 2b Solve for b, then solve for a. 2b + b = 12 3b = 12 b = 4 a = 2(4) = 8
The tens digit of a two-digit number is 2 more than the units digit. the number is 4 more than 6 times the sum of the digits. find the number.?
Call your number 10x + y. x = y + 2 and 10x + y = 4 + 6(x + y) Substitute y + 2 for x: 10(y + 2) + y = 4 + 6((y + 2) + y) This simplifies to 10y + 20 + y = 4 + 6y + 12 + 6y, ie 20 - 16 = 12y - 11y so y = 4 and x = 6 Your number is 64, which is indeed 4 more than the sum of its digits.
The units digit of a two digit number exceeds twice the tens digit by 1 what numbers if the sum of its digit is 10?
The units digit of a two digit number exceeds twice the tens digit by 1. Find the number if the sum of its digits is 10.
10a + b = 7(a + b) 10a + b = 7a + 7b 3a = 6b a = 2b (2,1), (4,2), (6,3), and (8,4) meet these requirements. So the numbers that are solutions are 21, 42, 63, and 84.
The number is 54. The sum of its digits is 5 + 4 = 9. 54/9 = 6.
(9,999) + (9,999) = 19,998 So it looks like the sum of two 4-digit numbers can't have more than 5 digits.
21 ways assuming no leading zeros, that is the smallest possible number with three digits is 100 Otherwise if leading zeros are allowed, it is 28 ways. . For three digit numbers the first digit can be 1-6. Leaving the remaining two digits to be 6-value_of_first_digit. For first digit 6, the… remain two digits sum to 0 which means they can only be 00, ie the number 600 - 1 number For first digit 5, the remain two digits sum to 1 which means they can be 01 or 10, ie the numbers 501, 510 - 2 numbers For first digit 4, the remain two digits sum to 2 which means they can be 02. 11, 20, ie the numbers 402, 411, 420 - 3 numbers etc It can be seen that the number of ways of making a two digit number (with leading zeros) sum to a number less than 10 is the required sum plus 1. So for the remaining first digits, the number of ways are: 3 -> 4 ways, 2 -> 5 ways, 1 -> 6 ways. Thus the total number of ways is 1 + 2 + .. + 6 = 21 If leading zeros are permitted, so that, for example, 060 (60) and 006 (6) are considered as three digits numbers, then there are a further 7 ways with a first digit of 0, making a total 28 ways. (MORE)
When you reverse the digits of your 2 digit number the new number is 45 less than the original number and the sum of the two digits times the difference of the two digits is 45 What are the two digit?
72 is the original number 72 - 27 = 45 (7+2)*(7-2) = 9*5 = 45 Here's how to do it via algebra: Choose x as ten's & y as the one's digit of your number. So your original number is 10x+y When you reverse the digits of the number the new number (10y+x) is 45 less than the original so: …10y+x =10x+y-45 10y-10y+x-x = 10x-x+y-10y-45+45 (rearrange) 9x-9y=45 (you can simplify by dividing by 9) x-y =5 [ 1st equation ] Next the sum of the digits times the difference of the digits is 45 so we have this equation: (x+y)*(x-y)=45 [ 2nd equation ] Now we have 2 equations with 2 unknowns. From the first equation : (x-y)= 5 we can substitute that value into the 2nd equation, so: (x+y)*(5)=45 (x+y) = 9 [ 3rd equation ] Now using the 3rd equation and the 1st equation we can solve for x & y. x-y= 5 x+y = 9 --------------- (add the 2 equations) 2x =14 x = 7 Substitute 7 into either equation above to solve for y. (7) + y=9 y = 9 - 7 = 2 So the tens digit is 7 & ones digit is 2 hence 72 is the original number. (MORE)
81 : 8 + 1 = 9, and 9 x 9 = 81. Here's how to figure: If the 2-digit number is ab,. then the number is equal to 10*a + b. And nine time the sum equals 9*(a + b). Set these two equal and you have: . 9*a + 9*b = 10*a + b --> 8*b = a. Knowing that the digits must be in the range 1 through 9: …1 x 8 = 8, and 2 x 8 = 16, so the only possibility is a=8 and b=1. (MORE)
Can you find three digits so that the sum of cubes of each digit gives a three digit number containing the same digits?
1 3 + 3 3 + 5 3 = 1 + 27 + 125 = 153. So the three digits are 1, 3 & 5.
What is the number of a two digit number that is four times the sum of its digits The tens digit is 3 less than the units digit what are you?
Let's break it down: "What is the two digit number ..." x is the 10's place; y is the 1's place; and z is the answer. 10x + y = z "... that is [equal to] 4 times the sum of its digits ..." 4(x + y) = 10x + y "... The tens digit is [equal to] 3 less than the units [1's] digit ..." x = y - 3 Brea…k down the complex equation: 4x + 4y = 10x + y 3y = 6x Substitute the other equation for x, because x = y - 3, and continue: 3y = 6(y - 3) 3y = 6y - 18 -3y = -18 -y = -6 y = 6 So 6 is y, the 1's number. x = y - 3 x = 6 - 3 x = 3 And 3 is x, your 10's number. Almost there! Now find your final answer, z: 10x + y = z 10(3) + 6 = z 30 + 6 = z 36 = z Your answer is 36. (MORE)
I did it by trial and error. 7^2 is 49. 4 + 9 = 13
The sum of the two digit number depends on what the number is for e.x. 27 + 32= 59. You get this by first adding the 7 and the 2 together, because if you would line the problem up and down the would be lined up directly below or above each other, getting an answer of nine. Then you add the 3 and the… 2 together getting an answer of 5. To complete your results you will put the five in the front because it is the sum of 3 and 2 and they were both in the front and then you will write the 9 at the end because it is the sum of the 7 and the 2 and they were both at the end of the number. (MORE)
Took me 15 mins but the answer is: 902 901 903 + 909 = 3615
1 + 6 = 7 x 3 = 21 5 + 4 = 9 x 3 = 27 6 + 3 = 9 x 3 = 27 4 + 9 = 13 x 3 = 39 2 + 5 = 7 x 3 = 21 3 + 8 = 11 x 3 = 33 None of the numbers listed are
If the digit '1' occupies the hundreds place then the other two digits total 8. 108, 117, 126, 135, 144, 153, 162, 171, 180 (9 three digit numbers) Similarly with '2 in the hundreds position the the remaining digits total 7. 207, 216, 225, 234, 243, 252, 261, 270 (8 three digit numbers) When the rem…aining numbers 3 to 9 are set as the lead digit then the number of three digit numbers formed decreases by one each time. The final total of such numbers is, 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45. (MORE)
I think that you would just add a .0 to the end of the two digit number, thus making it three digit, then divide. At the end, you have to put the decimal on the answer in the right place, though.
Observe that: 100+999 =1099, 101 + 998 = 1099 up to 549+550=1099. So 1099 * (1000-100/2) or 1099 * 450 (450 pairs of numbers that sum to 1099) or 494,500. This is a variant of the method exposed in the "Gauss Anecdote"
900 It is the only three digit number with a nine in the numbers place
The first digit must be 7 or greater, because the digits of 699 only sum to 24. Of the numbers beginning with 7, only 799 sums to 25, because if either of the other two digits are less than 9, the sum will be less than 25. That's one so far. For number beginning with 8, the other two digits must …sum to 17. That means they must be 9 and 8. So there are two suitable numbers beginning with 8, namely 889 and 898. That's a total of three so far. Finally, we have numbers beginning with 9. The other two digits must sum to 16, so the candidates are: 9 and 7, or 8 and 8. That gives us three numbers that fit the bill: 997, 979 and 988. That's six altogether. So the answer is six. (MORE)
33 is a two-digit number whose digits add up to 6. 60, 51, 42, 24, 15
How many three digit numbers are there such that sum of the squares of any two digits is equal to third digit?
18 101, 110, 112, 121, 125, 152, 211, 215, 228, 251, 282, 309, 390,512, 521, 822, 903, 930
What is the formula to find number of three digit numbers such that sum of the squares of any two digits is equal to third digit?
N squared would be used to find the square root of a number or numbers. In order to find the number of three digit numbers such that the sum of the square results of any two digits are equal to the third digit the use of the formula (HOE)squared=Hsquared*10000+2HE*100+Esquared is needed.
Almost all of the two digit composite numbers have two one digitfactors that have a sum.