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Z is defined as (K – L) ÷ 10

Expert replies
by abhi332 » Thu Feb 25, 2010 12:54 pm
K and L are each four-digit positive integers with thousands, hundreds, tens, and units digits defined as a, b, c,
and d, respectively, for the number K, and p, q, r, and s, respectively, for the number L. For numbers K and L, the
function W is defined as 5^a*2^b*7^c*3^d ÷ 5^p*2^q*7^r*3^s. The function Z is defined as (K – L) ÷ 10. If W = 16, what is the value
of Z?
(A) 16
(B) 20
(C) 25
(D) 40
(E) It cannot be determined from the information given.

[spoiler]OA:D[/spoiler]
Last edited by abhi332 on Thu Feb 25, 2010 2:22 pm, edited 1 time in total.
What you think, you become.
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Source: — Problem Solving |

by harsh.champ » Thu Feb 25, 2010 2:16 pm
abhi332 wrote:K and L are each four-digit positive integers with thousands, hundreds, tens, and units digits defined as a, b, c,
and d, respectively, for the number K, and p, q, r, and s, respectively, for the number L. For numbers K and L, the
function W is defined as 5a2b7c3d ÷ 5p2q7r3s. The function Z is defined as (K – L) ÷ 10. If W = 16, what is the value
of Z?
(A) 16
(B) 20
(C) 25
(D) 40
(E) It cannot be determined from the information given.

[spoiler]OA:D[/spoiler]
W is defined as 5a2b7c3d ÷ 5p2q7r3s.

This is a bit ambiguous.Are they multiplied or are they in the exponents??
Can you check with the question source ??
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by abhi332 » Thu Feb 25, 2010 2:22 pm
Corrected :)
What you think, you become.
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by florencejennifer » Thu Feb 25, 2010 10:33 pm
Given W= (5^a*2^b*7^c*3^d) ÷ (5^p*2^q*7^r*3^s)= 2^4

The Primes 5,7, and 3 can be ruled out if their power were equal.
Which implies a=p, c=r and d=s
i.e 2^b÷2^q = 2^4
2^(b-q) = 2^4
b-q = 4
Since b and q are Hundred digits values b-q=400

Hence K-L= a b c d - p q r s
= a b c d- a q c d
= 400

z= (K-L)÷10
= 400÷10
= 40
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by Anaira Mitch » Tue Dec 27, 2016 7:01 am
Given:
K = abcd = 1000a + 100b + 10c + d
L = pqrs = 1000p + 100q + 10r + s
W = 5a2b7c3d5a2b7c3d ÷ 5p2q7r3s5p2q7r3s = 5a−p2b−q3c−r5d−s5a−p2b−q3c−r5d−s = 16 = 2424

W can 16 only when W carries the powers of 2 only.
Hence b - q = 4 (i)
And the rest of the powers will be 0.
a= p, c = r, d = s (ii)

Required: Z = (K – L) ÷ 10 =?
Z = (abcd - pqrs)÷10 = (1000a + 100b + 10c + d) - (1000p + 100q + 10r + s) ÷ 10
Z = 1000 (a - p) + 100(b - q) + 10 (c - r) + 10 (d - s) ÷ 10
From equations (i) and (ii)
Z = 100(b-q) ÷ 10 = 100*4 ÷ 10= 40
Option D
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