Numbers PS problems- digits

[vishalwin]

Q1) Given a b c and d are distinct positive digits

AB + BA = CD4

which of the following can be value of A?

a) 1
b) 5
c) 7
d) 8
e) 9


Q2) Given a b and C are distinct positive digits

AB + BA = CDC

what can be the values of C and D?

[theCEO]

Q1) Given a b c and d are distinct positive digits

AB + BA = CD4

which of the following can be value of A?

a) 1
b) 5
c) 7
d) 8
e) 9

All letters are distinct.
The last digit of A+B is 4

Therefore
a) 1; a=1, b=3, 13+31=44; few digits so wrong
b) 5; a=5, b=9, 59+95=154; not all distinct values so wrong
c) 7; a=7, b=7, we can stop here as this is wrong, not all distinct
d) 8; a=8, b=6, 86+68=154; all distinct, correct
e) 9; a=9, b=5, this is the same as (b)so wrong

ans = d

[[email protected]]

Hi vishalwin,

The approach that theCEO used to answer your first question can also be used to answer your second question. There is a Number Property rule that you can use to your advantage to save some time though: when adding two 2-digit numbers together, if the total is a 3-digit number then the first digit of that number MUST be a 1.

AB + BA = CDC and A,B and C are DISTINCT positive digits....

AB + BA = 1D1

Now, we have B+A ends in 1...

Using that method, can you determine the possibilities?

GMAT assassins aren't born, they're made,
Rich

[Max@Math Revolution]

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


Q1) Given a b c and d are distinct positive digits

AB + BA = CD4

which of the following can be value of A?

a) 1
b) 5
c) 7
d) 8
e) 9

Let's consider only unit digit. Then we have only two cases, namely B+A=4 or B+A=14 (since A and B are digit numbers they are less than or equal to 9. That means B+A should be less than or equal to 18. So B+A cannot be 24 neither 34,... ). If B+A=4, since A+B is also 4, AB+BA=44(contradiction to the assumption). So B+A should be 14. That is AB+BA= (A+B)*10 + (B+A) = 154. That means C=1 and D=5.
Now let's find out what A and B are. Since A+B= 14, we have only tree cases to make 14, namely A+B= 7+7, 8+6 or 9+5. But A, B, C, D should be distinct. So A+B=8+6 (since the cases A=7, B=7 and B=5, D=5 have same digits). A can be, therefore, 8 or 6. So the answer is D.




Q2) Given a b and C are distinct positive digits

AB + BA = CDC

what can be the values of C and D?


Similarly with Q1, B+A=C or B+A=1C. If B+A=C, similarly with Q1, A+B=C ----> AB+BA=CC(contradiction to the assumption). So B+A=1C. That means AB+BA= (1C)*10 + 1C. So 1 is a carrying to the tenth digit. That means C=1 and D=C+1=2. So the answer is C=1, D=2.






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