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 Subject: "Multiplication of Consecutive No." Previous Topic | Next Topic
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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 thefollowing problem:If you multiply consecutive numbers from 1 to 15 what are the last 4numbers (from the right) of the answer.i.e. if1x2x3x4x5x6x7x8x9x10x11x12x13x14x15 = 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 beeither 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 astudent 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)    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, 3I 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)

Alert | IP Printer-friendly page | Edit | Reply | Reply With Quote | Top 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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