Say no to calculator!

Find the last 3 digits of 123456789 × 987654321. \large 123456789\times 987654321.

Hint: We do not need to multiply the 9 digit integers.


The answer is 269.

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9 solutions

Brodie Kushner
Mar 5, 2017

If you are attempting to find the last 3 digits - only the last three digits of each factor is required
So 789 X 321 =

1 X 789 = 789
20X 789 = 15,780 ( the 15,000 is irrelevant)
300 X 789 = 236,700 ( the 236,000 is irrelevant)

So - add the following 789 + 780 + 700 = 2269 - last three digits are 269

Moderator note:

This solution gives good motivation as to why we are caring about only the last 3 digits when applying the multiplication.

This technique can be generalized with the language of Modular Arithmetic .

This is basically how I solved it without having to multiply larger numbers. I took the first number away from the left side each time I multiplied and every time I added a zero to the end of the right.
1 X 789= 789
2 X 89= 780
3 X 9 = 700

Will U Ain't - 4 years, 3 months ago

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Yup, this is just long multiplication in disguise~~

Pi Han Goh - 4 years, 3 months ago

Could you explain why only last three digits of each factor are required? Does it mean that the answer would be the same if we multiply any pair of numbers ending with 789 and 321?

Pranshu Gaba - 4 years, 3 months ago

I used the exact same method as you!

Ishan Maheshwari - 4 years, 3 months ago

Do you have a trick for multiplying 20×789 and 300×789 quickly? I still had to whip out my calculator for that part.

Stacie Mahuna - 4 years, 3 months ago

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We can find out 789×2 and 789×3 first, and then append the required number of zeroes at the end. Do you know how to proceed now?

Pranshu Gaba - 4 years, 3 months ago

Ok guys, you are driving me nuts. This is the first time I hearing about Modular Arithmetic!! Very interesting, but my brain is going nuts! However, great thing to challenge it! Thank you

Adriana T. - 4 years, 2 months ago

Darn it. I got the right answer, but I could have done that better

Half-god Dragon - 3 years ago
Md Zuhair
Feb 28, 2017

This is Very simple, And it can be approached with Modular Arithmetic

We get 123456789 = 789 m o d 1000 123456789 = 789\mod1000 .

Again 987654321 = 321 m o d 1000 987654321 = 321\mod1000 .

Multiplying these two gives us

123456789 987654321 = 269 m o d 1000 123456789 * 987654321 = 269 \mod1000

So 269 \boxed{269} are the last three digits of the number.

Max Carvalho...I disagree. Most everybody else has solved it by recognizing that the last three digits of each factor determine the last three digits of the product. The use of modular arithmetic just puts that intuitive notion on rigorous mathematical footing.

Eric Lucas - 4 years, 3 months ago

Its quite vivid that 789 789 and 321 321 will be aiding to figure out the last three digits though workings seems to different :)

Naren Bhandari - 4 years, 3 months ago

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Ya true that

Md Zuhair - 4 years, 3 months ago

like that ;-)

Nafiz Imtiaz - 4 years, 3 months ago

Fairly sophisticated method for this one to be fair, haha. :)

Max Carvalho - 4 years, 3 months ago

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Probably True. How would you solve it?

Agnishom Chattopadhyay - 4 years, 3 months ago

Now i have to google modular arithmetic..

Kevin Murray - 4 years, 3 months ago

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Actually, that's not a bad idea. You'll learn something very useful. See modular arithmetic

It is essentially like Clocks. What do you get when you add 9 + 4 in a clock? You get 1: basically you get 12 and wrap back to 1 in some sense.

Agnishom Chattopadhyay - 4 years, 3 months ago

That is also my solution!

Luke Limbo - 4 years, 2 months ago
Naren Bhandari
Feb 27, 2017

To evalute the last three digits of 123456789 × 987654321 123456789 \times 987654321 we just need to focus on the 3-digit numbers 789 × 321 = 253269 789\times 321 = 253269 last digits are 269 \boxed{269}

Further more

The difference is 789 321 = 468 789-321 = 468 . Dividing 468 468 by 2 2 we get 234 234 and 789 × 321 789\times 321 can be written as

( 555 + 234 ) × ( 555 234 ) (555 + 234)\times (555-234)

= ( 555 ) 2 ( 234 ) 2 = (555)^2 - (234)^2

= 308025 54756 = 308025 - 54756

We require 3-digit number however 025 025 is smaller than 756 756 so better is to 4digit number 1025 which is nearest to 756 756 (which is not compulsory as we can take 2025,3025…8025)

= 1025 756 = 269 = 1025 - 756 = 269

So last three digit are 269 \boxed{269} .

Funny, in Excel it appears that the product is 121,932,631,112,635,000.000

Chuck Burks - 4 years, 3 months ago

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Really? Maybe there is a way to increase the size of the integers in Excel

Agnishom Chattopadhyay - 4 years, 3 months ago

In this case we only need to consider the product of last three digits of the two numbers 789 × 321 789 \times 321 because the rests of the numbers will not affect the last three digits of the product. Consider the long multiplication below.

789 × 321 789 1578 2367 269 \begin{array} {lr} & 789 \\ \times & 321 \\ \hline & 789 \\ & 1578 \ \ \\ & 2367 \ \ \ \ \\ \hline & \cdots 269 \\ \hline \end{array}

Therefore, the last three digits of the product is 289 \boxed{289} .

Alternatively, using modular arithmetic, we have:

789 × 321 789 × 1 + 89 × 20 + 9 × 300 (mod 1000) 789 + 1780 + 2700 (mod 1000) 789 + 780 + 700 (mod 1000) 2269 269 (mod 1000) \begin{aligned} 789 \times 321 & \equiv 789 \times 1 + 89 \times 20 + 9 \times 300 \text{ (mod 1000)} \\ & \equiv 789 + 1780 + 2700 \text{ (mod 1000)} \\ & \equiv 789 + 780 + 700 \text{ (mod 1000)} \\ & \equiv 2269 \equiv \boxed{269} \text{ (mod 1000)} \end{aligned}

Your solution is not obvious to follow.

How do you get 789 × 321 789 × 1 + 89 × 20 + 9 × 300 (mod 1000) 789 \times 321 \equiv 789 \times 1 + 89 \times 20 + 9 \times 300 \text{ (mod 1000)} ?

Obviously this congruence is true (if we did the manual tedious arithmetic), but it would be helpful to explain how you formed your congruence so that it allows others to better understand what you're trying to say.

I think you're trying to do some long multiplication in this step, am I right?

Pi Han Goh - 4 years, 3 months ago

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Yes, you are right. I will do some explanation. I was because it is difficult to present.

Chew-Seong Cheong - 4 years, 3 months ago
Jesse Nieminen
Mar 5, 2017

123456789 × 987654321 789 × 321 211 × 321 731 269 ( m o d 1000 ) 123456789 \times 987654321 \equiv 789 \times 321 \equiv -211 \times 321 \equiv -731 \equiv 269 \pmod{1000}

Hence, the answer is 269 \boxed{269} .

Right, but finding the last 3 digits of 211x321 is still relatively menial right?

Is there a way to skip this step? (I don't know how.... yet)

Pi Han Goh - 4 years, 3 months ago
Sutapa Saha
Mar 10, 2017

Guys it's​just simple trick. Remember how u multiply in lower classes So in that way 123456789 987654321


.............789 .............78 .............7


..............269

B D
Aug 31, 2018

You can't use a calculator. How I solved it is multiplying the last three digits 9 x 1 = 9(last digit) 8 x 2 = 16(don't write 16 only 6 and on the next one will add the 1)(middle digit) 7 x 3 = 21 + 1 = 22(don't write 22 only 2 and on the next one will add the 2) and get the answer 269.

Syed Shahabudeen
May 1, 2017

Vedic method for multiplication saved my day.

Viki Zeta
Mar 7, 2017

Other solutions there are enough to explain in an easy way. Here is the traditional one

123456789 × 987654321 = 81 81 × 123456789 × 987654321 = 123456789 × ( 987654321 × 81 ) 81 = 123456789 × 80000000001 81 = 121932631112635269 123456789 \times 987654321 \\ = \dfrac{81}{81} \times 123456789 \times 987654321 \\ = \dfrac{123456789 \times (987654321 \times 81)}{81} \\ = \dfrac{123456789 \times 80000000001}{81} \\ = 121932631112635269

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