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Atul Jayavant
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Dec0101, 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 
Dec0101, 06:13 AM (EST) 

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): 2·3 = 6 (mod 10) 6·6 = 6 (mod 10) 6·7 = 2 (mod 10) 2·8 = 6 (mod 10) 6·9 = 4 (mod 10) 4·11 = 4 (mod 10) 4·12 = 8 (mod 10) 8·13 = 4 (mod 10) 4·14 = 6 (mod 10) 6·3 = 8 (mod 10) 

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Judith Ross
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Jan0502, 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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