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Algebra

Blazing goIITian

Joined:	26 Mar 2009
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5 Aug 2009 20:21:24 IST

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882

Remainder problems:

None

Find the remainders when:

1. 7¹⁰³ is divided by 5

2. 32^{32power 32} is divided by 7

3. 32³² is divided by 3

4. 2222⁵⁵⁵⁵ + 5555²²²² is divided by 7

Challenging questions

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Comments (4)

swapnil joshi

Blazing goIITian

Joined: 10 Apr 2009

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5 Aug 2009 20:30:37 IST

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all solutions are in arihant algebra, its better if u see there

Aakash Anuj

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5 Aug 2009 20:35:27 IST

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If i could have, then i would not have posted the qstns here

AIR 538 Asish

Blazing goIITian

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5 Aug 2009 22:32:42 IST

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Q1. Observe the units place of various powers of 7 .....7^1 = 7, 7^2=9, 7^3=3, 7^4=1, 7^5=7 ... So 7^(4m) has units digit 1.... 7^(4m+1) has units digit 7 ... 7^(4m+2) has units digit 9 and 7^(4m+3) has units digit 3 ...................... 7^(103) has units digit 3 (as 103 is of the form 5m+3) ... So if u divide by 5, ull get the remainder as 3.

®µD®A

Blazing goIITian

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5 Aug 2009 23:05:05 IST

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These are quite easy if you know congruency...

7¹⁰³ is divided by 5

From Fermat Little Theorem,

$7^4equiv 1mod5Rightarrow 7^{103}equiv343equiv3mod5$

32^{32power 32} is divided by 7

$32^6equiv 1mod7Rightarrow 32^{32^{32}}equiv2^{32}equiv2^{30}cdot2^2equiv4mod7$

So leaves remainder 4 when divided by 7..

32³² is divided by 3

$32^2\equiv1\mod3\Rightarrow 32^{32}\equiv1\mod3$

2222⁵⁵⁵⁵ + 5555²²²² is divided by 7

$2222^6\equiv1\mod7\\\Rightarrow2222^{6\times925}\equiv1\mod7\\\Rightarrow 2222^{5555}\equiv2222^5\equiv3^5\equiv5\mod7$

$5555^6\equiv1\mod7\\\Rightarrow5555^{6\times370}\equiv1\mod7\\\Rightarrow 5555^{2222}\equiv5555^2\equiv4^2\equiv2\mod7$

So, 2222⁵⁵⁵⁵ + 5555²²²² leaves remainder 0 when divided by 7

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