In the multiplication problem above, F, G, and H represent unique odd

Question

123.jpg In the multiplication problem above, F, G, and H represent unique odd digits. What is the value of the three-digit number FGF? (A) 151 (B) 161 (C) 171 (D) 313 (E) 353

saswata4s · Answer

FGF
x G
----
HGG

Now, (C) and (E) are out because they provide 4 digits number after multiplication(17 x 7 = 119; 35 x 5 = 105).

(D) is out because for (D) : FGF = HGG; (313 x 1 = 313)

For (B)  : 161 x 6 = 966 (FGF x G = HGG - Satisfied)
 For (A)  : 151 x 5 = 755 (FGF x G = HGG - Satisfied)

Both A & B can be the answer.

mikemcgarry · Answer

Dear SajjadAhmad , I'm happy to respond. :-)

First, notice that F x G has to be G, or at least has to have a units digit of G. The three with F = 1 all work. For (D), it's true that F x G = G, but of course 313 x 1 = 313, and this doesn't fit the pattern. (D) is incorrect. For (E), F x G = 15, which has a units digit of G = 5. Notice, though, that 5 x 300 = 1500, so 5 x 353 > 1500, and this is NOT a three-digit number. (E) is incorrect. In fact, 17 x 7 = 119, so 170 x 7 = 1190, and 171 x 7 = 1197, also not a three-digit number. (C) is incorrect. That leaves us with (A) & (B). Both fit the digits pattern, but notice that the prompt specifies that F, G, and H are unique ODD digits. Thus, (B) is out, because one of its digits is the even number 6. That leaves (A). OA = (A) We never had to do the calculation, but it's not hard: 15 x 5 = 75 so 150 x 5 = 750 so 151 x 5 = 755 Does all this make sense? Mike :-)

Shiv2016 · Answer

Bunuel  and  mikemcgarry

I solved it using this approach. It would be great if you could give your insights on this.

F,G, and H are odd DIGITS. Therefore they can only take values 1,3,5,7, and 9.

F*G= G
That is possible in two cases:
(1) If both (F and G) are 1.
(2) If one of the two is 1 and that would be F.
(3) If both are 5 (5*5=5) but in this case we would not get 5 at tens place as there will be a carryover. (So cancelled)

Therefore D and E eliminated.

We are left with 151, 161, and 171.

FGF*G= HGG

F*G= G (no carryover) 
G*G= G 
Therefore G ends in 5 or 6 (because 5*5=ends in 5 and 6*6= ends in 6)
But we are told that the digits are even. Therefore 161 is eliminated. 
171 is not the answer because 7*7= ends in 9 and not 7.

The answer thus is 151.

mikemcgarry · Answer

Dear Shiv2016 , I'm happy to respond. :-)

I have some thoughts about how you started. You suggested F = G = 1 or F = G = 5, but the problem states quite clearly that all three numbers are unique , i.e. distinct. Thus, we cannot have repeats. Also, you assumed that F*G = G. You over looked the possibility that F*G is a two digit number with a last digit of G. For example, if G = 5, then F could be any odd number, and the unit digit of the product will be 5. As it happens, we get no answer choices of that sort, but given the stem, theoretically, the answer could have been in that form, and your approach would have missed it. My friend, does all this make sense? Mike :-)

AkshayKS21 · Answer

Question says digits are odd. But B has 6 which is even, so it is out. So A is the answer

Sajjad1994 · Answer

Official Explanation

At first it looks as though you’ll have to substitute every odd digit for the three variables until you stumble onto the correct answer, but there’s a trick to this problem that eliminates such guesswork. Look at the units column of the problem and you’ll see that F × G yields a product with a units digit of G; therefore, it is quite possible that F is

1. Actually, you can go even further: F must equal 1, because G cannot equal 1 (otherwise, the product of this multiplication problem would be FGF, not HGG). Plus, if both F and G were odd digits greater than 1, the product of this multiplication problem would be a four-digit number. Because the product is the three-digit number HGG, F equals 1. You can now eliminate (D) and (E), and you also know that G must equal 5 or 7 (G cannot equal 6 because the problem says G must be an odd digit). When G equals 7, the product is a four digit number; therefore, G is 5, and the   correct answer is (A)  . But you could have just plugged the answer choices into the problem, one at a time, to see which one works

Hope it Helps

ScottTargetTestPrep · Answer

Solution:

If F is not 1, then G must be 5. Since only 5 times an odd number will have the units digit 5. If F is 1, the G can be any odd digit (except 1 since F and G are unique). Let’s analyze the former case first.

If F is not 1, then G must be 5. However, since F and G are unique, F can’t be 5 either. We see that F is either 3 or 7. However, no matter if FGF is 353 or 753, when it’s multiplied by G = 5, the product will be a 4-digit number, not a 3-digit number. Therefore, we see that it is not true that F is not 1. In other words, F must be 1 and FGF is 131, 151 or 171. Let’s test each of these numbers.

If FGF is 131 and G is 3, we have 131 x 3 = 393. This can’t be HGG.

If FGF is 151 and G is 5, we have 151 x 5 = 755. This can be HGG. (Although we’ve found the answer, let’s check the last case to make sure the product is not in the form of HGG anyway.)

If FGF is 171 and G is 7, we have 171 x 7 = 1197. This can’t be HGG.

Answer: A

HoudaSR · Answer

By looking at the answer choices, we can see that F is either 1 or 3. Also G can only be 1,5 or 6 because these are the only numbers with cyclicity 1. G can't be 1 if F is 1 because these two are supposed to be distinct. and G can't be 6 because we are told it is odd. G is then definitely 5. 
From here we have only three possible answers for FGF: 151, 353 or 313
151x5= 755
353x5= 1765
313x1= 313
Hence FGF is 151

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In the multiplication problem above, F, G, and H represent unique odd

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