To find the remainder of ( 2^{33} ) when divided by 17, we can use Fermat's Little Theorem, which states that if ( p ) is a prime and ( a ) is an integer not divisible by ( p ), then ( a^{p-1} \equiv 1 \mod p ). Here, ( p = 17 ) and ( a = 2 ), so ( 2^{16} \equiv 1 \mod 17 ). Thus, ( 2^{33} = 2^{32} \cdot 2 \equiv (2^{16})^2 \cdot 2 \equiv 1^2 \cdot 2 \equiv 2 \mod 17 ). Therefore, the remainder is 2.
Preston in yellow jug or equiv. mixed 50/50 with distilled water
Let the number be ( x ). According to the problem, we have two conditions: ( x \equiv 2 \mod 4 ) and ( x \equiv 1 \mod 5 ). The first condition implies that ( x ) can be expressed as ( x = 4k + 2 ) for some integer ( k ). Substituting this into the second condition gives ( 4k + 2 \equiv 1 \mod 5 ), which simplifies to ( 4k \equiv -1 \equiv 4 \mod 5 ), leading to ( k \equiv 1 \mod 5 ). Thus, ( k = 5m + 1 ), and substituting back gives ( x = 4(5m + 1) + 2 = 20m + 6 ). The smallest positive solution occurs when ( m = 0 ), giving ( x = 6 ).
To find the least number that meets these conditions, we can express the problem in terms of congruences. We have: ( x \equiv 25 \mod 35 ) ( x \equiv 35 \mod 45 ) ( x \equiv 45 \mod 55 ) These congruences can be rewritten as: ( x \equiv -10 \mod 35 ) ( x \equiv -10 \mod 45 ) ( x \equiv -10 \mod 55 ) Since all three congruences have the same form, we can solve for ( x ) using the least common multiple of the moduli (35, 45, and 55), which is 1035. Adding 10 to account for the negative remainder, we find that ( x = 1035 - 10 = 1025 ). Thus, the least number is 1025.
12 oz, it measures on the troy scale.
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Lubricant typeHusqvarna or equiv. at 50:1 Right off the web site I use husky oil
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A square number can only yield specific remainders when divided by 3. When a number ( n ) is divided by 3, it can have a remainder of 0, 1, or 2. The possible square results are ( 0^2 \equiv 0 ), ( 1^2 \equiv 1 ), and ( 2^2 \equiv 1 ) (mod 3). Thus, the only possible remainders when dividing a square number by 3 are 0 or 1, never 2.
The 2010 Milan requires 5W20 Synthetic Blend (Motorcraft preferred or equiv.)
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