abc is a three-digit number in which a is the hundreds digit, b is the

Question

abc is a three-digit number in which a is the hundreds digit, b is the tens digit, and c is the units digit. Let \(&(abc)& = (2^a)(3^b)(5^c)\). For example, \(&(203)& = (2^2)(3^0)(5^3) = 500\). For how many three-digit numbers abc does the function &(abc)& yield a prime number?

(A) Zero
(B) One
(C) Two
(D) Three
(E) Nine

(A) Zero
(B) One
(C) Two
(D) Three
(E) Nine

abhinav770 · Answer

Great question! Is the answer (B), one ? Waiting for the OA.

CounterSniper · Answer

since 2 ,3,5 are fixed 
The required result is only possible when we have a 2 or a 3 or a 5 as the final value
The only possible value is 
100
for which we get 2 as the result

Imo B

please correct me if I am wrong .

EgmatQuantExpert · Answer

Solution 
  Given:  
 • abc is a three-digit number
•  &(abc)& = (2^a) * (3^b) * (5^c)

To find:  
 • The number of three-digit numbers, abc, for which &(abc)& yields a prime number

Approach and Working:   
 • For,  &(abc)& = (2^a) * (3^b) * (5^c) , to be a prime number, the possible cases are,
 o a = 1 and b = c = 0, or
o a = b = 0 and c = 1, or
o a = c = 0 and b =1
o In any other case, &abc& cannot be a prime number
o And, for abc to be a three-digit number, 'a' cannot be 0

Therefore, the only possible case is when a = 1 and b = c =0

Hence, the correct answer is option B.

Answer: B

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Free-Session-Algebra-e1533292089856.jpg

tatz · Answer

Good question.. Initially i mis-read the question. I thought we have to find three digit prime number being generated by taking different values of a, b, c. But the question is about yielding a prime number which may be of 1 digit / 2 digits / 3 digits etc.
 My logic-  Since the expression (2^a)(3^b)(5^c) contains 2, 3 & 5 in product form, the number getting generated out of any value of a, b, c will always be divisible by 2 or 3 or 5. So, that number cant be prime.

2 , 3 & 5 all are prime numbers. So, if we end up in finding a value of a, b. c such that it yields either 2, 3 or 5 then our purpose is served. We can do that in three ways
1) a= 0, b=1, c=0, this gives product of abc as 3 but "abc" becomes a two digit number (10) as a =0 (remember question asks for three digit value for abc)
2) a=1, b = 0, c= 0, this gives product of abc as 2 and also results in "abc"as a three digit number (100)
3) a=0, b=0, c=1, this gives product of abc as as 5, but "abc" becomes one digit number (1) as a =0, b=0
Any other value will not result in a prime number

Hence, a=1, b = 0, c= 0 satisfies both the requirements of three digit number (abc) and product a*b*c = prime number.
Answer=B

chetan2u · Answer

&(abc)& = (2^a)(3^b)(5^c)

We can say for sure that a is non zero digit, as abc is a 3-digit number.

(2^a)(3^b)(5^c)  will be prime only when the two powers are zero and the third is 1.

Since  (2^a)(3^b)(5^c)  is a prime and a is non zero, both b and c have to be zero, and a has to be 1.

&(abc)& = (2^a)(3^b)(5^c)=2^1*3^0*5^0=2

Only one possibility.

B

bumpbot · Answer

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abc is a three-digit number in which a is the hundreds digit, b is the

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abc is a three-digit number in which a is the hundreds digit, b is the

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