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Subject: "Multiplication of Consecutive No."     Previous Topic | Next Topic
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Conferences The CTK Exchange Middle school Topic #53
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Atul Jayavant
guest
Dec-01-01, 06:02 AM (EST)
 
"Multiplication of Consecutive No."
 
   Dear All,

My son aged 10 years, studying in India in Standard V came across the
following problem:

If you multiply consecutive numbers from 1 to 15 what are the last 4
numbers (from the right) of the answer.

i.e. if
1x2x3x4x5x6x7x8x9x10x11x12x13x14x15 = XXXXXXXXXXABCD, then what is A,B,C and D?

I tried to solve it as follows:

I picked up 2, 5, 10, 15 (i.e. 5x3) and 4 (2x2)

thus I have 2x5=10, 10 and 5x2=10, i.e. 3 tens which means that B,C and D will be 0,0 and 0.

re. 'A', as there are even numbers in the multiplication, 'A' will be
either 2,4,6 or 8 (but not 10, as there are no 5s remaining).
I am unable to go further to determine 'A'.

The answer is 8,0,0,0.

Please solve the problem, considering that the problem is put to a
student of 10 years.

Please reply to 'atuldjayavant@vipluggage.com'.

Regards,

Atul Jayavant


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alexb
Charter Member
672 posts
Dec-01-01, 06:13 AM (EST)
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1. "RE: Multiplication of Cons ..."
In response to message #0
 
   >1x2x3x4x5x6x7x8x9x10x11x12x13x14x15 = XXXXXXXXXXABCD, then
>what is A,B,C and D?
>
>I tried to solve it as follows:
>
>I picked up 2, 5, 10, 15 (i.e. 5x3) and 4 (2x2)
>
>thus I have 2x5=10, 10 and 5x2=10, i.e. 3 tens which means
>that B,C and D will be 0,0 and 0.

So, by now, you have used the factors 5, 10, 5 (from 15) and 4 (twice 2). What remains is to detemine the last digit of the product of the following numbers:

2, 3, 6, 7, 8, 9, 11, 12, 13, 14, 3

I am going to multiply them one after another, but instead of keeping the whole product shall only write the last digit (this is what "mod 10" means):

23 = 6 (mod 10)
66 = 6 (mod 10)
67 = 2 (mod 10)
28 = 6 (mod 10)
69 = 4 (mod 10)
411 = 4 (mod 10)
412 = 8 (mod 10)
813 = 4 (mod 10)
414 = 6 (mod 10)
63 = 8 (mod 10)


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Judith Ross
guest
Jan-05-02, 07:21 AM (EST)
 
2. "RE: Multiplication of Cons ..."
In response to message #1
 
   A variation on the above solution could be to break each of the remaining numbers
2, 3, 6, 7, 8, 9, 11, 12, 13, 14, 3 into its prime factors:

2, 3, 2*3, 7, 2*2*2, 3*3, 11, 2*3*3, 13, 2*7, 3

and then find the last digit for the appropriate power of each prime.

There are 7 factors of 2, so look at the pattern for the last digit of the powers of 2:

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32 (ends in 2 and the pattern 2, 4, 8, 6 repeats)
Therefore 2^7 ends in 8.

Similar patterns exist for the other primes. Finally multiply the last digits as in the previous solution.


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