This is how I solved.
f(abc) = 3 * f(def)
So, 5^a*2^b*3^c = 3*[5^d*2^e*3^f]
So, (5^a*2^b*3^c)/(5^d*2^e*3^f) = 3
So, 5^(a-d)*2^(b-e)*3^(c-f) = 3^1
So,5^(a-d)*2^(b-e)*3^(c-f) = 3^1*5^0*2^0
S0, a-d =0, b-e =0 and c -f =1.
So, a =d, b = e and c = f +1
So abc - def is equal to abc - abf
And since c = f + 1, differrence is 1
Answer is A

Since f(abc) = 3*f(def), I would assume that f = c - 1 from the function above.

The answer should be (A).
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Incognito,

Could you please detail your work?

Thanks

f(abc)/f(def)=5^(a-d) 2^(b-e) 3^(c-f) = 3^1 -> a = d, b = e, c = f+1 -> A is the answer

Sure. The function consists purely of powers of 5, 2 and 3. None of which are multiples of each other.

Since f(abc) = 3*f(def), and since the function is entirely made up of multiples of 5, 2 and 3, it would imply that the only difference is a power of 3. This would mean that a = d (since there is no multiple of 5 from the functions output), b = e (no multiple of 2 in the functions output) and c = f + 1 (since there is a multiple of 3 in the two functions output). Since abc and def are 3-digit numbers, and c = f + 1, the difference between abc and def would be 1. Hope that helps!
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f(abc)=3*f(def) so the only difference is one more multiple of 3 in f(abc). The only way for this to happens is if z is one more. z is the units digit so it's total value is 1 more.

Answer A

I went totally tangent on this one.

thought abc is a*b*c and not a number abc

and wasnt sure why every one was thinking its too obvious.

lesson learnt!

Similar questions to practice:
the-function-f-is-defined-for-each-positive-three-digit-100847.html
k-and-l-are-each-four-digit-positive-integers-with-thousands-91004.html
the-three-digit-positive-integer-x-has-the-hundreds-163241.html
for-any-four-digit-number-abcd-abcd-3-a-5-b-7-c-11-d-126522.html
k-and-l-are-each-four-digit-positive-integers-with-thousands-126646.html

Hope it helps.
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oh yeah..tricky one
f(def) = 5^d*2*e*3^f
a is the same multiplied by 3
we have same prime factorization
abc = def+1
abc-def = 1

https://gmatclub.com/forum/the-function ... 00847.html

The answer is "A" - VeritasPrepKarishma gives an incredibly clear explanation in a similar problem linked above.

f(abc) = 5^a * 2^b *3^c
f(def) = 5^d * 2^e * 3^f

since, f(abc) = 3*f(def)
5^a * 2^b *3^c = 5^d * 2^e * 3^(f+1)

so, a=d
b=e
c=f+1

so, for example, if abcis 123, then def is 124,

hence difference is 1.

My issue was even getting to this step:
So, 5^a*2^b*3^c = 3*[5^d*2^e*3^f]

How are we to conclude that abc = def = xyz?

I never even got off the ground on this one.

f(xyz) = \(5^x 2^y 3^z\)

=> f(abc)=\(5^a 2^b 3^c\)

=> f(def)=\(5^d 2^e 3^f\)

=> f(abc)=\(5^a 2^b 3^c\) = 3 * f(def)=\(5^d 2^e 3^f\)

=> \(\frac{[5^a 2^b 3^c]}{ [5^d 2^e 3^f]}\) = 3

=> \(5^{a - d} 2^{b - e} 3^{e - f}\) = 3

=> \(5^{a - d} 2^{b - e} 3^{e - f} = 5^0 2^0 3^1\)

Comparing the bases:

=> a - d = 0
=> b - e = 0
=> c - f = 1

Adding all:

=> (a + b + c) - (d + e + f) = 0 + 0 + 1

Answer A
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