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06 Jan 2022, 04:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
81% (01:08) correct
19%
(01:23)
wrong
based on 54
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a12a is a four-digit number with the digit a in both the thousand’s and unit’s places. What is the value of a; if the number is divisible by 6?
(A) 2
(B) 3
(C) 4
(D) 6
(E) 8
(A) 2
(B) 3
(C) 4
(D) 6
(E) 8
08 Jan 2022, 13:01
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A number is divisible by 6 if its also divisible by 2 AND 3.
A number is divisible by 2 if it's even and by 3 if the sum of digits is divisible by 3.
Answer is D
A number is divisible by 2 if it's even and by 3 if the sum of digits is divisible by 3.
Answer is D
08 Jan 2022, 13:43
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Fun question:
a12a is a four-digit number with the digit a in both the thousand’s and unit’s places. What is the value of a; if the number is divisible by 6?
Going through the four steps as I would on the test:
What is given?
A four-digit number: A12A (it's important to note that A's are DIGITS in a number, and not variables being MULTIPLIED).
I'm also told (as part of the question), that this number should be DIVISIBLE BY 6.
I have a suspicion about what I can do with that.
STEP 2: WHAT IS ASKED
Pretty straightforward question here: what is 'a'. I notice, though, it's what IS the value of A and not what COULD BE the value of A. At first this kind of surprised me, as I thought there would be a chance a had multiple possible values. Time to work through using the process I thought of in step 1.
STEP 3: PLAN and STRATEGIZE
Well, again, I kind of started doing this in step 1. Knowing that A12A was divisible by 6 means that A must be even: 0,2,4,6, or 8. 0 doesn't work, though, because that would not be a four-digit number! (0s before the first non-zero digit don't count. 0s after the first digit do).
So A is either 2,4,6,or 8. B is already gone.
Knowing that A12A is divisible by 6 *also* means that A+1+2+A = 2A + 3 => must be a multiple of 3 (Multiples of 6 must be multiples of 3, and the digits of multiples of 3 add up to multiples of 3).
STEP 4: SOLVE
So which A makes 2A + 3 a multiple of 3? I could plug in the answers and solve (except 3, already eliminated). It's not terribly slow. But if I *really* know my GMAT number properties I know that if 2A + 3 = Mult3, then 2A must be a mult3, and so A must be a mult3.
Only 6 works.
(If I did plug in: 2(2) + 3 = 7, 2(4)+3 = 11, 2(6)+3 = 15, 2(8) + 3 = 19. only a=6 works)
D.
a12a is a four-digit number with the digit a in both the thousand’s and unit’s places. What is the value of a; if the number is divisible by 6?
Going through the four steps as I would on the test:
What is given?
A four-digit number: A12A (it's important to note that A's are DIGITS in a number, and not variables being MULTIPLIED).
I'm also told (as part of the question), that this number should be DIVISIBLE BY 6.
I have a suspicion about what I can do with that.
STEP 2: WHAT IS ASKED
Pretty straightforward question here: what is 'a'. I notice, though, it's what IS the value of A and not what COULD BE the value of A. At first this kind of surprised me, as I thought there would be a chance a had multiple possible values. Time to work through using the process I thought of in step 1.
STEP 3: PLAN and STRATEGIZE
Well, again, I kind of started doing this in step 1. Knowing that A12A was divisible by 6 means that A must be even: 0,2,4,6, or 8. 0 doesn't work, though, because that would not be a four-digit number! (0s before the first non-zero digit don't count. 0s after the first digit do).
So A is either 2,4,6,or 8. B is already gone.
Knowing that A12A is divisible by 6 *also* means that A+1+2+A = 2A + 3 => must be a multiple of 3 (Multiples of 6 must be multiples of 3, and the digits of multiples of 3 add up to multiples of 3).
STEP 4: SOLVE
So which A makes 2A + 3 a multiple of 3? I could plug in the answers and solve (except 3, already eliminated). It's not terribly slow. But if I *really* know my GMAT number properties I know that if 2A + 3 = Mult3, then 2A must be a mult3, and so A must be a mult3.
Only 6 works.
(If I did plug in: 2(2) + 3 = 7, 2(4)+3 = 11, 2(6)+3 = 15, 2(8) + 3 = 19. only a=6 works)
D.
