For all positive integers j and k, the operation k j is defined to b

Question

For all positive integers  j  and  k , the operation k ◇ j is defined to be the product of  j  consecutive integers, beginning with  k . For example, 6 ◇ 4 = 6 x 7 x 8 x 9. If  a  = 20 ◇ 20 and b = 21 ◇ 20, then  \frac{a}{b}  =

(A)  \frac{21}{40}

(B)  \frac{20}{39}

(C)  \frac{20}{21}

(D)  \frac{1}{5}

(E)  \frac{1}{2}

Bunuel · Answer

a = 20 ◇ 20 = 20 * 21 * ... * 39  (the product of 20 consecutive integers, beginning with 20)

b = 21 ◇ 20 = 21 * 22 * ... * 40  (the product of 20 consecutive integers, beginning with 21)

Therefore:

\frac{a}{b}=

=\frac{20 * 21 * ... * 39}{21 * 22 * ... * 40}=

=\frac{20}{ 40}=

=\frac{1}{ 2}

Answer: E.

DanTheGMATMan · Answer

Visualize a and b as clipped off factorials, then set up a/b:

https://www.youtube.com/watch?v=jzz2_fiEuCk

Heix · Answer

couldn't understand why you stop upto 39 and 40 can you please explain this

Bunuel · Answer

You're multiplying 20 consecutive integers starting from 20, so the last number will be 20 + 20 - 1 = 39. Similarly, for b, you start at 21 and multiply 20 consecutive numbers, ending at 21 + 20 - 1 = 40. That's why a goes from 20 to 39 and b from 21 to 40.

gmatuser523 · Answer

pro tip - always remember 0 --> 9 is 10 numbers - this one always confuses me but it logically makes sense: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are 10 numbers

That way 10, 11, 12, 13, 14 ,15, 16, 17, 18, 19 are 10 numbers

so the 20 numbers in 20 (diamond) 20 would be:

20 * 21 * (...) * 29 * 30 * 31 * (...) 39

which comes out to 20 numbers

Same applies for 21 square 20, except this time you shift one number down. Then from there you divide & cancel out to get 20/40 = 1/2

For all positive integers j and k, the operation k j is defined to b

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For all positive integers j and k, the operation k j is defined to b

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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