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n m, with n and m

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n m, with n and m

by sanju09 » Tue May 18, 2010 5:09 am
If D is a two digit number (so D = n m, with n and m digits) what is the last digit m of D?
(1) The number 3 D is a three digit number whose last digit is m.
(2) The digit m is less than 7.
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by thinkpanther » Tue May 18, 2010 10:11 am
What is the OA?

IMO E


Only 1 -

n = 3 m = 5

nm =35 and 3nm = 105

m could be 5

Also n =6 m =0
nm = 60 and 3nm = 180

m could be 0

so inconclusive

Only 2 -

Could be 0-6

Both 1 and 2 -

Same as 1

hence E
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by sk818020 » Tue May 18, 2010 7:33 pm
Here is another way to think about it.

D is a two digit number n (in the tens place), m (in the units place). Thus,

D=10n+m

(1) Says that 3 D is a 3 digit number with m in the units place.

3D = 3(10n+m) = 30n+3m

You only need to worry about 3m because the units place of 3m will be the same as 3D.

So m must be a one digit number that when multiplied by 3 has the same number in the digits spot.

5 and 0 are the only one digit number that when multiplied by 3 the result has the same unit digit.

5*3=15.
0*3=0.

(2) Does not help us because it m could be equal to any number between 0-6.

Putting them together we are still left with m=0,5.

Both are not sufficient, E.
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by liferocks » Tue May 18, 2010 7:38 pm
From1
Units digit of 3m is m
..this is possible when m=0 or 5..not sufficient

From 2
m<7...not sufficient

combining, condition 2 does not add any new information to condition 1..not sufficient

Ans option E
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