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In the multiplication problem above, F, G, and H represent unique odd
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14 Mar 2017, 10:10
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In the multiplication problem below, F, G, and H represent unique odd digits. What is the value of the threedigit number FGF? (A) 151 (B) 161 (C) 171 (D) 313 (E) 353
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Re: In the multiplication problem above, F, G, and H represent unique odd
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14 Mar 2017, 12:02
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.
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Re: In the multiplication problem above, F, G, and H represent unique odd
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14 Mar 2017, 12:11
SajjadAhmad wrote: In the multiplication problem below, F, G, and H represent unique odd digits. What is the value of the threedigit number FGF?
(A) 151 (B) 161 (C) 171 (D) 313 (E) 353 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 threedigit number. (E) is incorrect. In fact, 17 x 7 = 119, so 170 x 7 = 1190, and 171 x 7 = 1197, also not a threedigit 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
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Re: In the multiplication problem above, F, G, and H represent unique odd
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03 Apr 2017, 04:26
Bunuel and mikemcgarryI 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.
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Re: In the multiplication problem above, F, G, and H represent unique odd
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15 May 2017, 13:28
Shiv2016 wrote: Bunuel and mikemcgarryI 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. 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
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Re: In the multiplication problem above, F, G, and H represent unique odd
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17 Jun 2017, 05:24
saswata4s wrote: 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. Question says digits are odd. But B has 6 which is even, so it is out. So A is the answer



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Re: In the multiplication problem above, F, G, and H represent unique odd
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21 Feb 2019, 11:25
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 fourdigit number. Because the product is the threedigit 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
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Re: In the multiplication problem above, F, G, and H represent unique odd
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