Not so difficult

Number Theory Level 2

Find the last two digits of the number 7 100 − 3 100 7^ { 100 } - 3^ { 100 }


The answer is 0.

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Shivam Jadhav by Shivam Jadhav , 22, India

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2 solutions

Shivam Jadhav
Mar 14, 2015

Use euler's theorem .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

which eulers formula

Harshi Singh - 6 years, 1 month ago

Wow ! You are really a lucky fellow to have gotten an upvote for this awesome solution .

A Former Brilliant Member - 6 years ago
Parth Bhardwaj
Mar 14, 2015

first calculate last two digits of 7^100 - last two digits of powers of 7 are - 01,49,43,01,07,49,43,01,07.......... and so on we find that last two digits are 01 when power of 7 is a multiple of 4. Thus last wo digits of 7^100 are 01 (100 is a multiple of 4)
last two digits of power of three are 03,09,27,81,43,29.87,61,83,49...............
But we dont find any pattern in it. 3^100 can also be written as (3^10)(3^10)(3^10)(3^10)(3^10)(3^10)(3^10)(3^10)(3^10)(3^10). And we find that last digits of all these brackets are 49. So answer would be the last two digits of 49^10 which is 01 ( we can find that by observing a pattern in the last two digits of powers of 49 - 49,01,49,01,49,......)
So our final answer would be 01-01 = 00


0 Helpful 0 Interesting 0 Brilliant 0 Confused

Why did you ask for "last two digits"? When you just gave '0' in the last? I knew 0 wud be the last digit. This is not fair!

Kartikay Kaul - 6 years, 3 months ago

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