If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2i3k5m7p, then i + k + m + p =
A) 4
B) 7
C) 8
D) 11
E) 12
GRE Question of the Day: If x is the product of the positive integers from 1 to 8
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well i think that since the possible no’s can be 1,2,3,4,5,6,7,8 n then its given x=2i3k5m7p then 2,3,5,7 are already there so the remainin no’s r 1,4,6,8 so the ans will be 1+4+6+8=19
so its none of the above…..
its D)
x=2^7*3^2*5*7
d
its D
D
no, i did not get it… can someone explain a bit more simply?
D)
Hey aparna,
Break down each number from 1 to 8 into its prime factors(basically obtain the product of 1 to 8 in terms of 2,3,5 and 7)
1= 1
2=2 = 2^1
3=3
4=2×2 = 2^2
5=5
6=3X2 = 3^1 x 2^1
7=7
8= 2x2x2 = 2^3
u get the product of numbers from 1to 8 as (2^7)x(3^2)x(5^1)x(7^1)
so u get i=7, k=2, m=1,p=1.
i+k+m+p = 11
ans is 11
i=7
k=2
m=1
p=1
d
d
ans is 11
11
ya{d}
x = 1*2*3*4*5*6*7*8
= 1*2*3*(2*2)*5*(2*3)*7*(2*2*2)
= 2^7*3^2*5*7
i+k+m+p = 11
ans. D
as x is product of positve integers 1to 8 and from the question it is clear that only 1,4,6,8 are missing .hence 1+4+6+8=19
so NONE OF THE ABOVE
D
hey guys this is simple
x is produt of 1,2,3,4,5,6,7,8
x=1*2*3*4*5*6*7*8
1*2*3*(2*2)*5*(2*3)*7*(2*2*2)
(2*2*2*2*2*2*2)(3*3)*5*7
2^7*3^2*5*7
11
so ans is (d)11