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]
Z is defined as (K – L) ÷ 10
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- harsh.champ
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W is defined as 5a2b7c3d ÷ 5p2q7r3s.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]
This is a bit ambiguous.Are they multiplied or are they in the exponents??
Can you check with the question source ??
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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
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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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
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