Prime factorize 96:
96 = (2^5)*3
Thus, 96 has only two odd divisors: 1 and 3.
If m is odd, then the units digit of m is odd. If the units digit is odd, the units digit must therefore be either 1 or 3, but it cannot be 1: we can't find two single-digit numbers which multiply to 96. So if the units digit is odd, the units digit must be 3 (and the number could be 843 or 483).
Knowing that the hundreds digit is 8 is insufficient; the number could be 843 or 834, or 862 or 826.
Digits: The product of the units digit...
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II
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Thanks Ian.
What made you start the question by factorizing 96 and go down that route. What was your thinking behind this ... what was the trigger in the question which sent you down this route.
when I tackled this question, I didnt even think about factoring 96 ... I basically approached it as follows.
Q-Stem:
m = HTU (where H=hundreds digit, T=Tens digit, U=Units digit)
H*T*U = 96
Question is asking us to find U (the units digit of m)
(1) m is odd.
This means that the units digit (U) could be either 1,3,5,7, or 9 so that m is odd.
We know that H*T*U = 96
Lets run through the values 1,3,5,7, and 9 to see if they are factors of 96. By applying the div rules we can quickly eradicate some of the values:
96 doesnt end in a 5 or 0 ... so it cannot be div by 5. So we can cross this out.
96. 9+6 = 15 ... this is not div by 9. cross out 9.
we have 1,3, and 7 left.
there is no div shortcut for 7. but we can quickly find out that 96 is not div by 7. So cross out.
left with 1 and 3.
96/1 = 96 ... so this mean H*T have to equal 96. the largest number H*T can be is 9*9 which equals 81. So we from this we know U cannot be 1. Which means it has to be 3.
96/3 = 32.
32=8*4
This means Units digit is 3. SUFF.
(2) huindreds digit of m is 8
96/8 = 12
What are the (single digit) factors of 12: 1,2,3,4,6
So which two numbers make 12 (2*6, 3*4, 4*3, 6*2).
From this we can see that the units digit could be 6, 4, 3, or 2. INSUFF.
Hence answer A.
What made you start the question by factorizing 96 and go down that route. What was your thinking behind this ... what was the trigger in the question which sent you down this route.
when I tackled this question, I didnt even think about factoring 96 ... I basically approached it as follows.
Q-Stem:
m = HTU (where H=hundreds digit, T=Tens digit, U=Units digit)
H*T*U = 96
Question is asking us to find U (the units digit of m)
(1) m is odd.
This means that the units digit (U) could be either 1,3,5,7, or 9 so that m is odd.
We know that H*T*U = 96
Lets run through the values 1,3,5,7, and 9 to see if they are factors of 96. By applying the div rules we can quickly eradicate some of the values:
96 doesnt end in a 5 or 0 ... so it cannot be div by 5. So we can cross this out.
96. 9+6 = 15 ... this is not div by 9. cross out 9.
we have 1,3, and 7 left.
there is no div shortcut for 7. but we can quickly find out that 96 is not div by 7. So cross out.
left with 1 and 3.
96/1 = 96 ... so this mean H*T have to equal 96. the largest number H*T can be is 9*9 which equals 81. So we from this we know U cannot be 1. Which means it has to be 3.
96/3 = 32.
32=8*4
This means Units digit is 3. SUFF.
(2) huindreds digit of m is 8
96/8 = 12
What are the (single digit) factors of 12: 1,2,3,4,6
So which two numbers make 12 (2*6, 3*4, 4*3, 6*2).
From this we can see that the units digit could be 6, 4, 3, or 2. INSUFF.
Hence answer A.
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Ian Stewart
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When I see that the question is about a 'product' of positive integers, and that I will need to identify the divisors, I know that this will be easiest to do if I prime factorize. It isn't strictly necessary here, and your approach is good, though by prime factorizing, notice that you save yourself the trouble of trying to divide 96 by 7. Still, either method should let you comfortably answer the question in less than two minutes.II wrote:Thanks Ian.
What made you start the question by factorizing 96 and go down that route. What was your thinking behind this ... what was the trigger in the question which sent you down this route.
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rhymes_with_luck
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