Integer problem 4

What are the last 4 digits of 7¹⁰⁰?

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Solution 1

Calculating the last 4 digits of 7⁴, 7⁸, 7¹⁶, ...

7²=49.
7³=343.
7⁴=2401.
From now on, only the last 4 digits are calculated:
The last 4 digits of 7⁸=7⁴*7⁴ are 4801.
The last 4 digits of 7¹⁶=7⁸*7⁸ are 9601.
The last 4 digits of 7²⁰=7¹⁶*7⁴ are 2001.
The last 4 digits of 7⁴⁰=7²⁰*7²⁰ are 4001.
The last 4 digits of 7⁸⁰=7⁴⁰*7⁴⁰ are 8001.
And the last 4 digits of 7¹⁰⁰=7⁸⁰*7²⁰ are to be 0001.

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Solution 2

Utilizing 7²=49

  7¹⁰⁰
=(50-1)⁵⁰
=50⁵⁰-...-(50*49*48/6)*50³+(50*49/2)*50²-50*50+1.
Since all the terms except for the last 3 terms are multiples of 10000, letting 'n' be an integer,
  7¹⁰⁰
=10000n+(50*49/2)*50²-50*50+1
=10000n+3062500-2500+1
=10000n+3060001.
Hence, the last 4 digits of 7¹⁰⁰ are to be 0001.

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Solution 3

Utilizing 7⁴=2401

  7¹⁰⁰
=(2400+1)²⁵
=2400²⁵+...+(25*24/2)*2400²+25*2400+1.
Since all the terms except for the last 2 terms are multiples of 10000, letting 'n' be an integer,
7¹⁰⁰=10000n+60001.
Hence, the last 4 digits of 7¹⁰⁰ are to be 0001.

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Mathematical Problems

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© 2006 Takashi Shimazaki
Updated: April 14, 2013