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**SOME SAMPLE QUESTIONS**

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**Divisibility and Congruence**

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**Case 1: What is the remainder when each of the numbers 1, 7, 16, 91, is divided by 3? The answer is obviously 1.**

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**Case 2: What is the remainder when the number 3, 27, 83, 131 are divided by 8? The answer to this question is 3. Let us define the congruence of numbers. **

**Two numbers a and b are congruent with respect to m if both a and b leave same remainder when divided by m.**

Mathematically, a ≡ b(mod m)

Important point in the congruence of numbers is that in problems involving divisibility of numbers, a number may always be replaced by another number congruent to the first.

Mathematically, a ≡ b(mod m)

Important point in the congruence of numbers is that in problems involving divisibility of numbers, a number may always be replaced by another number congruent to the first.

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**For example, 91 ≡ 1 (mod 3). Therefore 91 may be replaced by 1. Let us learn to use this concept.**

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**Problem 1: Prove that 2 ^{20} – 1 is divisible by 41.**

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**Sol: We know that 2 ^{20 }– 1 =_{ }(2^{5})^{4} – 1 = (32)^{4} – 1**

Let us check number congruent to 32 with respect 41.

-9, 32, 73, 114, 155, 196, etc. are all congruent to 32.

There we may replace 32 by any of these congruent numbers.

Since -9 is smallest (numerically), therefore replacing 32 by -9, we get

(–9)^{4} – 1^{ }= (81)^{2} – 1

Now, numbers congruent to 81 with respect to 41 are -1, 40, 81, 122, 163 etc.

Replacing 81 by -1, we get

(-1)^{2 }– 1 = 0

Thus 2^{20} – 1 leaves remainder zero when divided by 41

Hence 2^{20} – 1 is divisible by 41

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**Problem 2: Prove that (53) ^{103} (103)^{53} is divisible by 39.**

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**Sol: 53 is congruent to 14 and 103 is congruent to -14 with respect to 39, therefore replacing 53 and 103 by 14 and -14 respectively, we get (14) ^{103} (-14)^{53} = (14)^{103} – (14)^{53} = 14^{53} (14^{50} - 1) Since 14^{53} and 39 are co-prime, therefore 14^{53} cannot be multiple of 39. So let us consider 14^{50} – 1 only 14^{50}- 1 = (14^{2})^{25} – 1 = (196)^{25} – 1 196 is congruent to 1 with respect to 39, therefore replacing 196 by 1, we get (1)^{25} – 1 = 0 (remainder when 14^{50} – 1 is divided by 39) Thus 14^{50} – 1 is divisible by 39. Hence (53)^{103} (103)^{53} is divisible by 39**

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