If 7 divides $32^{32^{32}}$ then the remainder is ___

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Solution

We have 32 = $2^{5}$
$∴ (32)^{32} = (2^{5})^{32} = 2^{160} (32)^{32} = (3 - 1)^{160} =^{160} C_{0} 3^{160} -^{160} C_{1} 3^{159} + \dots -^{160} C_{159} 3 +^{160} C_{160} I = 3 (3^{159} -^{160} C_{1} 3^{158} + \dots -^{160} C_{159}) + I = 3 m + 1, m ϵ I_{+} N o w, 32^{32^{32}} = 32^{3 m + 1} = 2^{5 (3 m + 1)} = 2^{15 m + 5}$
$∴ 32^{32^{32}} = 2^{3 (5 m + 1)} {.2}^{2} = 4. (8)^{5 m + 1} = 4. (7 + 1)^{5 m + 1} = 4. (^{5 m + 1} C_{0} (7)^{5 m + 1} +^{5 m + 1} C_{1} (7)^{5 m} +^{5 m + 1} C_{2} (7)^{5 m - 1} + \dots +^{5 m + 1} C_{5 m} 7 +^{5 m + 1} C_{5 m + 1})$
$= 4 [7 {^{5 m + 1} C_{0} 7^{5 m}} +^{5 m + 1} C_{1} 7^{5 m - 1} +^{5 m + 1} C_{2} 7^{5 m - 2} + \dots +^{5 m + 1} C_{5 m} + 1] = 4 [7 n + 1], n ϵ I_{+}$
= 28n + 4
This show that when $32^{32^{32}}$ is divided by 7, then the remainder is 4.

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