Find the last 3 digits of 1 2 3 4 5 6 7 8 9 × 9 8 7 6 5 4 3 2 1 .
Hint: We do not need to multiply the 9 digit integers.
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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
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Yup, this is just long multiplication in disguise~~
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?
I used the exact same method as you!
Do you have a trick for multiplying 20×789 and 300×789 quickly? I still had to whip out my calculator for that part.
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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?
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
Darn it. I got the right answer, but I could have done that better
This is Very simple, And it can be approached with Modular Arithmetic
We get 1 2 3 4 5 6 7 8 9 = 7 8 9 m o d 1 0 0 0 .
Again 9 8 7 6 5 4 3 2 1 = 3 2 1 m o d 1 0 0 0 .
Multiplying these two gives us
1 2 3 4 5 6 7 8 9 ∗ 9 8 7 6 5 4 3 2 1 = 2 6 9 m o d 1 0 0 0
So 2 6 9 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.
Its quite vivid that 7 8 9 and 3 2 1 will be aiding to figure out the last three digits though workings seems to different :)
like that ;-)
Fairly sophisticated method for this one to be fair, haha. :)
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Probably True. How would you solve it?
Now i have to google modular arithmetic..
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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.
That is also my solution!
To evalute the last three digits of 1 2 3 4 5 6 7 8 9 × 9 8 7 6 5 4 3 2 1 we just need to focus on the 3-digit numbers 7 8 9 × 3 2 1 = 2 5 3 2 6 9 last digits are 2 6 9
Further more
The difference is 7 8 9 − 3 2 1 = 4 6 8 . Dividing 4 6 8 by 2 we get 2 3 4 and 7 8 9 × 3 2 1 can be written as
( 5 5 5 + 2 3 4 ) × ( 5 5 5 − 2 3 4 )
= ( 5 5 5 ) 2 − ( 2 3 4 ) 2
= 3 0 8 0 2 5 − 5 4 7 5 6
We require 3-digit number however 0 2 5 is smaller than 7 5 6 so better is to 4digit number 1025 which is nearest to 7 5 6 (which is not compulsory as we can take 2025,3025…8025)
= 1 0 2 5 − 7 5 6 = 2 6 9
So last three digit are 2 6 9 .
Funny, in Excel it appears that the product is 121,932,631,112,635,000.000
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Really? Maybe there is a way to increase the size of the integers in Excel
In this case we only need to consider the product of last three digits of the two numbers 7 8 9 × 3 2 1 because the rests of the numbers will not affect the last three digits of the product. Consider the long multiplication below.
× 7 8 9 3 2 1 7 8 9 1 5 7 8 2 3 6 7 ⋯ 2 6 9
Therefore, the last three digits of the product is 2 8 9 .
Alternatively, using modular arithmetic, we have:
7 8 9 × 3 2 1 ≡ 7 8 9 × 1 + 8 9 × 2 0 + 9 × 3 0 0 (mod 1000) ≡ 7 8 9 + 1 7 8 0 + 2 7 0 0 (mod 1000) ≡ 7 8 9 + 7 8 0 + 7 0 0 (mod 1000) ≡ 2 2 6 9 ≡ 2 6 9 (mod 1000)
Your solution is not obvious to follow.
How do you get 7 8 9 × 3 2 1 ≡ 7 8 9 × 1 + 8 9 × 2 0 + 9 × 3 0 0 (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?
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Yes, you are right. I will do some explanation. I was because it is difficult to present.
1 2 3 4 5 6 7 8 9 × 9 8 7 6 5 4 3 2 1 ≡ 7 8 9 × 3 2 1 ≡ − 2 1 1 × 3 2 1 ≡ − 7 3 1 ≡ 2 6 9 ( m o d 1 0 0 0 )
Hence, the answer is 2 6 9 .
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)
Guys it'sjust simple trick. Remember how u multiply in lower classes So in that way 123456789 987654321
.............789 .............78 .............7
..............269
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.
Vedic method for multiplication saved my day.
Other solutions there are enough to explain in an easy way. Here is the traditional one
1 2 3 4 5 6 7 8 9 × 9 8 7 6 5 4 3 2 1 = 8 1 8 1 × 1 2 3 4 5 6 7 8 9 × 9 8 7 6 5 4 3 2 1 = 8 1 1 2 3 4 5 6 7 8 9 × ( 9 8 7 6 5 4 3 2 1 × 8 1 ) = 8 1 1 2 3 4 5 6 7 8 9 × 8 0 0 0 0 0 0 0 0 0 1 = 1 2 1 9 3 2 6 3 1 1 1 2 6 3 5 2 6 9
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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