Waste less time on Facebook — follow Brilliant.
×

Amazing problem!

Hello friends, We have to find the last two digits of \(1\times3\times5\times7\times9....\times99\), I did and came out with 25.But I want to verify.

Note by Kishan K
4 years, 5 months ago

No vote yet
11 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

From the fact that it is divisible by \(25\) but not by \(2\), we can deduce that it ends with either \(25\) or \(75\). Now, consider the multiplications modulo \(4\): \( 1 \times 3 \times 5 \times 7 \times 9 \times \dots \times 99 \equiv 1 \times 3 \times 1 \times 3 \times \dots \times 3 \equiv 1^{25} \times 3^{25} \equiv 1^{25} \times (-1)^{25} \equiv -1 \equiv 75 \not\equiv 25 \), so the answer is \(\boxed{75}\).

Tim Vermeulen - 4 years, 5 months ago

Log in to reply

Argh, you beat me to it xD Well done.

Sotiri Komissopoulos - 4 years, 5 months ago

Log in to reply

how these type of shortcuts can be found? I want to know .if in any website it is plz tell me.

Budha Chaitanya - 4 years, 5 months ago

Log in to reply

I did not use any website. In fact, I never solved such a problem before. I guess you'll just need to be familiar with prime factors and the modulo operator.

Tim Vermeulen - 4 years, 5 months ago

Log in to reply

refer elemantary number theory by David M Burton

Naga Teja - 4 years, 5 months ago

Log in to reply

I think you used Chinese remainder theorem.If not then please tell me the complete theorem.

Kishan K - 4 years, 5 months ago

Log in to reply

He did not use Chinese Remainder Theorem. Which part of his solution are you confused with?

Sotiri Komissopoulos - 4 years, 5 months ago

Log in to reply

@Sotiri Komissopoulos I am confused over the multiplication modulo 4...

Krishna Jha - 4 years, 5 months ago

Log in to reply

@Krishna Jha That is simply the individual remainders left when \(4\) divides the given odd numbers which will obviously be \(3\) or \(1\) and you can easily find out the number of such terms which yield \(1\) or \(3\) as remainder.

Aditya Parson - 4 years, 5 months ago

Log in to reply

@Aditya Parson why only with 4, why not other?

Budha Chaitanya - 4 years, 5 months ago

Log in to reply

@Budha Chaitanya Because \(4\) divides \(100\), and looking at the last two digits of a number is essentially considering de remainder when that number is divided by \(100\). As \(4\) divides \(100\), and we know that the number is one less than a multiple of \(4\), then the last two digits of the number are also one less than a multiple of \(4\). Hence, it's a nice trick to use here, as \(75\) is one less than a multiple of \(4\), and \(25\) is not.

Tim Vermeulen - 4 years, 5 months ago

Log in to reply

@Tim Vermeulen thank you

Budha Chaitanya - 4 years, 5 months ago

Log in to reply

Its 75... courtesy mathematica ... :-P...Bt how do we solve these sort of problems is still a mystery to me..

Krishna Jha - 4 years, 5 months ago

Log in to reply

There is no theorem that says "oh, if you multiply the first 50 odd digits then the last 2 digits are 75." Rather, it takes intuition along with a knowledge of mathematical theorems. In this case, somebody figured "okay, well, there are more than 1 factors of "5" in this, but no factors of 2 so it must end in either 25 or 75. Then, to figure out if it was a 75 or 25 at the end, he counted the numbers of factors of 3, because any number ending in 25 when multiplied by 3 ends with 75, and any number ending in 75 multiplied by 3 ends with 25.

Michael Tong - 4 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...