Hi everyone,
I really don’t understand this question at all! I’ve checked the solution and the explanation (attached also) but still can’t wrap my head around it. I think the function element is confusing me and making it hard for me to see it clearly.
Can anyone explain this to me like I’m 5 years old? 
Thanks in advance 
P.S. I don’t tend to have issues with divisibility and primes so I think its the structure of the question that’s throwing me off.
Okay, to me, the first thing that needs to be clarified is where that weird 500 comes from, well, it is because 4x1x125=500 
So as the question stated above, function &(abc)&=(2^a)(3^b)(5^c). We are looking for a prime result in this multiplication. Among 2,3 and 5, the only three prime numbers are 3,5, or2, and 2 is a big no-no for prime result in multiplication unless it is only 2 itself ( in this scenario, we have(2^1)(3^0)(5^0)—>abc=100.) In other cases, as I said, in order to get a prime number, we have to discard 2 by trurning a to zero, but by doing so we could never form a three-digit number abc, so the only suitable case is 100.
Hope this messy explanation helps.
Ok so i’ll try to do my best in trying to explain this :3
If you didn’t understand the weird function thingy; here are two more examples:
&(155)& = (2^1)(3^5)(5^5) = 2x243x3125 ( too big of an example lol)
&(721)& = (2^7)(3^2)(5^1) = 128x9x5 = 5760
The units of the three digit number we have ( abc ), are the powers of 2,3 and 5.
The hundrets digit (a) being the exponent for 2,
the tenths digit ( b ) being the exponent for 3 and
the units digit ( c ) being the exponent for 5.
Now, i think you should focus on the (2^a) (3^b)(5^c) part only.
What nubers do a, b and c need to be for the multiplication, (2^a) (3^b)(5^c) to equal a prime? Notice how 2,3 and 5 are all primes.
So we need only one out of 2,3 and 5 for that multiplication to be a prime, and the rest needs to be 1. And for the rest to be equal to 1, they need to be in the power of 0:
(2^1) (3^0)(5^0) = 2x1x1 = 2 – > here; a=1 b=0 c=0 ; abc = 100
(2^0) (3^1)(5^0) = 1x3x1 = 3 – > here ; a=0 b=1 c=0; abc = 010
(2^0) (3^0)(5^1) = 1x1x5 = 5 – > here ; a=0 b=0 c=1; abc = 001
And like in the explanation given for the question, abc is a three digit leaving only 100, one option.
Hope i didnt make any calculation errors lol 
@Alice This is how I approached this problem.
Given, abc is a three digit number such that:
&(abc)& = (2^a)(3^b)(5^c) meaning The function will spit out a number that will contain 2,3,5 as its factor and the power of the prime number numbers 2,3,5 will be determined by the digits of abc.
What we need to find:
We need to find a three digit number that when substituted in the function will spit out a prime number.
Prime numbers have only two multiples. The number itself and one. Multiplying a prime number with another prime will not give us a prime number because it will have more than two multiple. For ex : 2 and 3 multiplied together give us 6. 6 has more than one multiple.
Now, keeping the above info in mind we can solve this question.
We cannot have a power greater than 0 for two digits in (2^a)(3^b)(5^c) at the same time because that would mean multiplying two prime together leading to a composite number.
So the only combinations that will work here are : 100, 010,001.
Only 100 is a three digit number. Hence there is only one three digit number that when substituted in the function will yeild a prime number.
Hope this helps. 
Thank you, @sisi ! I understand! For so long I was thinking that we had focus on what prime would be produced from the function and I couldn’t see how that was 100, I didn’t realise that it was just the format of “abc” that we needed. Thank you! 
Thank you, @Nilly.inthewoods and @divya! Your explanations are both so helpful I was so fixed on thinking that we had to focus on what prime was being produced from the function like 500. I was so confused because I kept thinking “100 isn’t a prime, how did they get that!?”, I didn’t fully understand that we were looking at “abc” as a number with hundredths, tenths, units digits, but now I see that. Your explanations both made that very clear to me, thank you so much.
