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Solution: No. The perfect cubes mod 7 are 0, 1, and 6. Now, every power of 2 is congruent to 1, 2, or 4 mod 7. Thus 2y + 15 ≡ 2, 3, or 5 mod 7. This is an impossibility. 187 Example Prove that 2k − 5, k = 0, 1, 2, . . never leaves remainder 1 when divided by 7. 51 Congruences Solution: 21 ≡ 2, 22 ≡ 4, 23 ≡ 1 mod 7, and this cycle of three repeats. Thus 2k − 5 can leave only remainders 3, 4, or 6 upon division by 7. 188 Example (A IME 1994) The increasing sequence 3, 15, 24, 48, . . , consists of those positive multiples of 3 that are one less than a perfect square.

This is because we can pair the 2n consecutive integers {a + 1, a + 2, a + 3, . . , a + 2n} into the n pairs {a + 1, a + n + 1}, {a + 2, a + n + 2}, . . , {a + n, a + 2n}, and if n + 1 integers are chosen from this, there must be two that belong to the same group. So now group the one hundred integers as follows: {1, 2, . . 20}, {21, 22, . . , 40}, {41, 42, . . , 60}, {61, 62, . . , 80} and {81, 82, . . , 100}. Pigeonhole Principle 25 If we select fifty five integers, we must perforce choose eleven from some group.

These equalities give (1), (2) and (3). Property (4) follows by successive application of (3), and (5) follows from (4). ❑ Congruences mod 9 can sometimes be used to check multiplications. For example 875961 · 2753 = 2410520633. For if this were true then (8+7+5+9+6+1)(2+7+5+3) ≡ 2+4+1+0+5+2+0+6+3+3 mod 9. But this says that 0 · 8 ≡ 8 mod 9, which is patently false. 177 Example Find the remainder when 61987 is divided by 37. Solution: 62 ≡ −1 mod 37. Thus 61987 ≡ 6 · 61986 ≡ 6(62)993 ≡ 6(−1)993 ≡ −6 ≡ 31 mod 37.