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Re: Multiplication the only way? [#permalink]
14 Dec 2010, 18:57

2*3*5*7*11*13*17*19

quick calculation:

3*10*77*13*17*19

231*10*13*17*19

231*10*221*19

231*221*190

Above steps did not take me much time, but after this I think one needs to be strong in multiplication!

If there is some short cut, it would be great to know....but I won't take the risk of some short cut which might result in wrong answer during the exam.

Re: Multiplication the only way? [#permalink]
14 Dec 2010, 19:15

Expert's post

My response from another thread:

Quickly approximate 2, 3, 5, 7, 11, 13, 17, 19 Make groups 2*5 = 10 3*17 = 50 (approximately) 7*13 = 100 (approximately) 11*19 = 200 (approximately) So you make 7 zeroes (the 2 and the 5 also make a 0). When you multiply all these, the answer will be close to 10^7 _________________

Re: Multiplication the only way? [#permalink]
15 Dec 2010, 05:50

"My response from another thread:

Quickly approximate 2, 3, 5, 7, 11, 13, 17, 19 Make groups 2*5 = 10 3*17 = 50 (approximately) 7*13 = 100 (approximately) 11*19 = 200 (approximately) So you make 7 zeroes (the 2 and the 5 also make a 0). When you multiply all these, the answer will be close to 10^7 "

Dear Karishma, a quick question on this: when i actually multiply the numbers, i get 10^6..there are only 6 zeros from what I see. What am I overlooking ? Thank you in advance!

Re: Multiplication the only way? [#permalink]
15 Dec 2010, 06:05

Expert's post

0987654312 wrote:

"My response from another thread:

Quickly approximate 2, 3, 5, 7, 11, 13, 17, 19 Make groups 2*5 = 10 3*17 = 50 (approximately) 7*13 = 100 (approximately) 11*19 = 200 (approximately) So you make 7 zeroes (the 2 and the 5 also make a 0). When you multiply all these, the answer will be close to 10^7 "

Dear Karishma, a quick question on this: when i actually multiply the numbers, i get 10^6..there are only 6 zeros from what I see. What am I overlooking ? Thank you in advance!

The 5 of 50 and the 2 of 200 above multiply to make another 0. _________________

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