INTEGERS

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INTEGERS

by bobdylan » Tue Jun 12, 2012 3:48 am
If 4a8 and 6b are integers with unspecified digits a and b, which of the following could be the result of multiplying 4a8 X 6b?
a. 20,574
b23,816
c.30,686
d. 32,507
e. 36,728

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by gmat_and_me » Tue Jun 12, 2012 5:19 am
Anything multiplied by 8 can not result in 7 in the units digit.
Hence eliminate D. The total value of the product has to be more
than 24000 since the most significant digits of the numbers are
4 and 6. Hence eliminate A and B. That leaves us with C and E.
To have 6 in the units place of the result, we can have 2 and 7,
while to have 8 we can have 1 or 6 in the units place of 6b. If
you divide E with 61 or 66 most significant digit will be 5 and
hence E can not be the answer.

HTH
bobdylan wrote:If 4a8 and 6b are integers with unspecified digits a and b, which of the following could be the result of multiplying 4a8 X 6b?
a. 20,574
b23,816
c.30,686
d. 32,507
e. 36,728

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by gmatwar13 » Tue Jun 12, 2012 8:02 am
Answer would be C...
How...At any case the digit will be greater that 24000... in that case a,b eliminated.
Similarly eliminate e... as at any case it can't be greater than 33xxx. left with c and d.. clearly d cant be answer as last digit can't be odd

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by Fighton!! » Tue Jun 12, 2012 8:43 am
since 4a8 ends with even no 4a8*6b will always be even eliminate D.

consider minimum and maximum value of 4a8*6b

Minimum (4a8*6b): 400*60=24000
Maxium(4a8*6b): 499*69 ( 500*70=35000) < 35000

so the value of 4a8*6b lies between (24000 and 35000)

24000--------<----------4a8*6b---------<------35000

eliminate A , B ,E

C left

answer is C