digit a of x

[sanju09]

If x is a two digit number (so that x = b a, with b and a as digits), what is the units' digit a of x?

(1)The number 3 x is a three digit number whose units' digit is a.

(2)The digit a is less than 7.

[albatross86]

There is a conceptual way to do this problem which is difficult to explain, but it is simply by observing the possibilities of a that satisfy this condition in your head. That should be easy for some, but anyway here's an algebraic explanation.

The number x can be expressed as 10b + a

1. 3*(10b + a)
= 30b + 3a

The units digit is therefore "3a" but truncated to its own units digit

You can confirm this with an example: 24*3 = 72 ... 4*3 = 12... so units digit of 24*3 = 2 which is the truncated units digit of 12, i.e. 12 - 10

Another example 98*3 = 294 ... 8*3 = 24 so units digit of 98*3 = 4 which is truncated units digit of 24 i.e. 24 - 20

Sometimes you won''t need to truncate, like 22 * 3 = 66, where the units digit is simply 2*3 = 6

So, either 3a = a => a = 0
OR 3a - 10 = a => a = 5
OR 3a - 20 = a => a = 10 ... but a is a digit so it cannot be greater than 9.

So a could be 0 or 5.

INSUFFICIENT

2. a < 7

Could be anything from 0 to 6

INSUFFICIENT

Both 1 and 2:

Could be 0 or 5.

INSUFFICIENT

Pick E.

[mj78ind]

So my approach here is to pick numbers and disprove ....

stmt 1 - say x = 31, 41 it works but x could also be 72, 82 ....... hence stmt 1 does not suffice

stmt 2 - does not suffice as shown above.

Combined not enough

Hence E

[albatross86]

So my approach here is to pick numbers and disprove ....

stmt 1 - say x = 31, 41 it works but x could also be 72, 82 ....... hence stmt 1 does not suffice

stmt 2 - does not suffice as shown above.

Combined not enough

Hence E

Hi, could you please explain how those examples you've given for statement 1 work?

As far as I can see, only numbers with 0 or 5 in the units digit would work.

[selango]

I am not sure whether my approach is correct.

Let x=10b+a

From stmt1,

3x is a 3 digits number(let pqr)

100p+10q+r=3x=30b+3a

100p+10q+r=30b+3a

r=a

-->100p+10q+a=30b+3a

2a=100p+10q-30b

a=50p+5q-15b

a/5=10p+q-3b

The equation will be satisified only if a is divisible by 5.Since a is a digit it can be 0 or 5 only.

a=0 or 5

Insufficient.

From stmt2,

a can be any no from 0 to 6.Insufficient.

Combining,

Again a can be 0 or 5.

Insufficient.

Hence E

[gmatmachoman]

If x is a two digit number (so that x = b a, with b and a as digits), what is the units' digit a of x?

(1)The number 3 x is a three digit number whose units' digit is a.

(2)The digit a is less than 7.

Unit digits range from 0 -9

St 1:

Multiply the unit digit(a) by 3 and see which one exactly matches with" a" in its unit's place.

a * 3= a( expected)
0 *3 = 0 (YES)
1*3= 3(NO)
2*3=6 (NO)
3*3=9 (NO)
4*3=2 (NO)
5*3=5 YES
6*3=8 (NO)
7*3 =1(NO)
8*3 =4 (NO)
9*3 =7 (NO)


a can be 0 or 5.. Insifficient

ST 2: again no much of any use.

combining..Not really of any use

Sp pick E

[mj78ind]

So my approach here is to pick numbers and disprove ....

stmt 1 - say x = 31, 41 it works but x could also be 72, 82 ....... hence stmt 1 does not suffice

stmt 2 - does not suffice as shown above.

Combined not enough

Hence E

Hi, could you please explain how those examples you've given for statement 1 work?

As far as I can see, only numbers with 0 or 5 in the units digit would work.

@abhay

My bad I was trying to make the hundred's digit of 3x same as units of x. And when I reread I saw my error.

Thanks