Can anyone help me or actually tell me the steps to finding the last 1/2/3 digits of a big number like this \(5^{287^{543}}\)

I need it badly........

No vote yet

2 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest5^(287^543) mod 1000

Note that the last 3 digits of 5^3, 5^4, 5^5, 5^6 are 125, 625, 125, 625

This suggests that 5^(odd number>1) mod 1000 = 125, and 5^(even number>2) mod 1000 = 625

You can prove the above by induction.

Since 287^543 is a multiplication of odd numbers, then it must be an odd number as well.

Thus 5^(287^543) mod 1000 = 5^(odd number>1) mod 1000 = 125

Log in to reply

Thanks for the simple sol. But what I want to know is that is there any way out other than using modulo arithmetic and for a general case? Like \(\6^{336^{775}}\) ?

Log in to reply

There's no escaping modulo arithmetic :)

It is obvious that 6^(336^775) is a multiple of 8, to find the last three digits, it suffices to find what it is congruent to modulo 125

By binomial theorem,

(5+1)^M is congruent to 1 + 5M + 25 MC2 modulo 125

It suffices to find the congruence of M modulo 25 and MC2 modulo 5

For your example: M = 336^775

MC2 = (M)(M-1)/2

and (M -1) is obviously a multiple of 5, thus MC2 is a multiple of 5

M is congruent to 11^775 modulo 25

11^5 is congruent to 1 modulo 25.

M is congruent to 1 modulo 25.

(5+1)^M is congruent to 1 + 5M + 25 MC2 modulo 125

which is congruent to 6 modulo 125

and is thus congruent to 256 modulo 1000

Also, the original question - to find 5^(287^543) modulo 1000 is even simpler. As this number is obviously a multiple of 125, it suffices to find what it is congruent to modulo 8

5^2 = 25 congruent to 1 mod 8, thus 5^(287^543) is congruent to 5 mod 8

the only multiple of 125 less than 1000 congruent to 5 mod 8 is 125, so we are done

You can solve in general by solving for the congruence mod 125 and mod 8. 5 and 6 are easier since they give a lot of free information, but this approach can be generalized reasonably easily.

Log in to reply

Log in to reply

Work out mod 10, mod 100, mod 1000 respectively.

Log in to reply

Actually i don't know how to do that. I'd be grateful 2 u if u could explain a bit further.

Log in to reply

Modulo Arithmetic

Log in to reply

Easy way to do it? Keep multiplying until you find a pattern.

Log in to reply