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No. Equiv is not allowed in Scrabble.

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What is the remainder when 2 to the power 33 is divided by 17?

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.


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When a number is divided by 4 the remainder is 2. When the same number is divided by 5 the remainder is 1. What is that number?

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 ).


The least number which when divided by 35 leaves remainder of 25 when divided with 45 leaves a remainder of 35 and when divided by 55 leaves 45 as remainder is?

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.


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