The sum is 22 times the sum of the three digits.

There is no number. 100 is the only one that has a hundreds digit. It's impossible.

Mental math like if 12 into 72 it will not be a double digit if you divide 100 by 10 it will be double digits. Most times if you have a tripple digit number and divide it by double digit the answer will be a double digit.

The first digit has to be in the set [1-6] because, if it were zero, it would not be a three digit number. The number of combinations, not using duplicates, of three numbers from the set [0-6] is 6 times 6 times 5, or 180.

38977 is in ones place9 is in tens place (and is three times the number in thousands place)8 is in the hundreds place3 is in the thousands place7+9+8+3=27

what is the greets possible 9 digit number that uses each of the digits 1-3 times

To count the number of times a digit occurs in an integer, start by initializing an array of ten counts of digits, such as int digits[10];Then, in a loop while the number is non zero, increment the element in the digits array that corresponds to the units digit, and then divide the number by ten, such as digits[number%10]++ and number/=10;int digits[10];int i;int number = some number;for (i=0; ido (digits[number%10]++; number/=10) while (number != 0);

They are the same because they are both multiplication. They also can be the same if the two digit number times by the one digit number equals a three digit number. They are different because the 3 digits number will obviously produce a higher product.

15 1 x 5 = 5, and 5 x 3 = 15

1 digit number: only 1 number 2 digits number: 18 numbers 3 digits number: 76 So there are 95 numbers containing 9.

The answer to this question is at my website aim feliciano980

18. The sum of its digits is 9. And yes, it is unique.

It can only be a multipule of 6. 12.

i am a two digit # my tens digit# is 3 times my ones digit #and the sum of my digit is 12 what am i

Let's assume there are 5 total digits, although that would have been handy information to include. All we can deduce from the available information is 4 of the digits. 318_2 There are ten possible digits for the 4th number, leaving 10 possible numbers.

77....724 where the 7 is repeated as many times as you like, making the number [possibly] infinite. However, I suspect the question requires a three digit number consisting of the digits of the set {2, 4, 7} used once each: 724

None, the digits are the same.

A decimal is a way of representing a number in such a way that the place value of each digit is ten times that of the digit to its right. A decimal representation does not require a decimal point. It can have any number of digits: from one to infinitely many.

Call the digits 'A' and 'B', so that the number is written: AB.10A + B = 4 (A+B)10A + B = 4A + 4B6A = 3B2A = B ---> any 2-digit number where the units digit is twice the tens digit12 or 24 or 36 or 48

Assuming you're allowed to repeat numbers, the answer is 81. For each digit, you have three choices: 2, 5, and 6. So there are three one-digit numbers you can make, and each one digit number can be combined with any of three other digits to make a two-digit number, giving 9 (3 times 3) possibilities. Now you can take any of the nine two-digit numbers as the tens and ones places in the four-digit number and any other two-digit number as the thousands and hundreds places, which gives 81 (9 times 9) possibilities. If you can't repeat numbers, the answer is zero because you don't have enough numbers to fill all the digits.

Any three digit number which is in the four times table. E.g 100, 104, 108 ... 992, 996.

Three hundred thirty-three million, two hundred twenty-two thousand, one hundred eleven.

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