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# the function h(n) is defined to be product of all the

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10 Jan 2006, 22:34
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the function h(n) is defined to be product of all the integers from 2 to 8, inclusive. h=2^n * 3^m * 5^k * 7^l
n+m+k+l=?
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10 Jan 2006, 22:39
2*3*4*5*6*7*8 = 2 * 3 * 2^2 * 5 * 2 * 3 * 7 * 2^3 = 2^7 * 3^2 * 5 * 7

So n = 7, m = 2, k = 1, l = 1

n+m+k+l = 11
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10 Jan 2006, 22:57
11.

Equated the two values and combined prime factors.
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10 Jan 2006, 23:32
2* 3* 4 * 5 * 6 * 7 * 8 = 2^7 * 3^2 * 5 ^1 * 7^1

so n+m+k+l = 7+2+1+1 = 11
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11 Jan 2006, 06:30
laxieqv wrote:
the function h(n) is defined to be product of all the integers from 2 to 8, inclusive. h=2^n * 3^m * 5^k * 7^l
n+m+k+l=?

2^n * 3^m * 5^k * 7^l = 2x3x4x5x6x7x8x
2^n * 3^m * 5^k * 7^l = 2x3x2x2x5x2x3x7x2x2x2
2^n * 3^m * 5^k * 7^l = 2^7 x 3^2 x 5 x 7

n+m+k+l=7+2+1+1
n+m+k+l=11
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11 Jan 2006, 07:13
i must be missing something. what are m, k and l ?
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12 Jan 2006, 19:51
Found the Prime factors of 2*3*4*5*6*7*8

2^n * 3^m * 5^k * 7^l = 2x3x4x5x6x7x8x
2^n * 3^m * 5^k * 7^l = 2x3x2x2x5x2x3x7x2x2x2
2^n * 3^m * 5^k * 7^l = 2^7 x 3^2 x 5 ^1x 7^1

upon equating powers,

n+m+k+l=7+2+1+1

n+m+k+l=11
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12 Jan 2006, 21:46
ywilfred wrote:
2*3*4*5*6*7*8 = 2 * 3 * 2^2 * 5 * 2 * 3 * 7 * 2^3 = 2^7 * 3^2 * 5 * 7

So n = 7, m = 2, k = 1, l = 1

n+m+k+l = 11

Great illustartion Wilfred.

Agreed it`s 11.
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13 Jan 2006, 07:01
good job, buddies~!
OA is 11.
13 Jan 2006, 07:01
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