ALPHAMETIC QUESTIONS ON THE GMAT

Question

ALPHAMETIC QUESTIONS ON THE GMAT 
  AN INTRODUCTION TO SOLVING ALPHAMETIC QUESTIONS ON THE GMAT  
  BY KARISHMA, VERITAS PREP

Today, let’s learn how to solve alphametics. An alphametic is a mathematical puzzle where every letter stands for a digit from 0 – 9. The mapping of letters to numbers is one-to-one; that is, the same letter always stands for the same digit, and the same digit is always represented by the same letter.

First focus on the big picture of the alphametic – such as, a two number is added to another two digit number to give a three digit number etc. Then look at the nitty gritty – for which digit can each letter stand?

Question 1:  With # and & each representing different digits in the problem below, the difference between #&& and ## is 667. What is the value of &?

https://www.veritasprep.com/blog/wp-content/uploads/2015/04/Alphametic0.jpg

(A) 3
(B) 4
(C) 5
(D) 8
(E) 9

Solution:   The big picture: A two digit number is subtracted from a three digit number to give 667. So the three digit number must be a bit larger than 667. This means that the hundreds digit of #&& must be either 6 or 7. It cannot be 8 because you cannot obtain 800+ by adding a two digit number to 667.

Let’s look at both cases:

# is 6: If you subtract 66 from 6&&, you will not get 667 – the largest value you can get is 699 – 66 = 633. So # cannot be 6.

# must be 7.

Now the question is very simple

7&& – 77 = 667

7&& = 667 + 77 = 744

Answer (B)   This question is discussed   HERE

There are many other ways in which you can solve this question including plugging in the answer choices. We should now take a look at a DS question on alphametics.

Question 2:

https://www.veritasprep.com/blog/wp-content/uploads/2015/04/Alphametic.jpg

In the correctly worked addition problem above, M, N, P, R, S, T and V are distinct digits. Is R > 3?

Statement 1: M, N and P are positive even integers.

Statement 2: S = 2

Solution:   This is certainly harder than the PS question but our process will remain the same.

First, let’s see what information we are given in the question – the units digits of all three numbers are the same. The two-digit numbers add up to give a three digit number. The hundreds digit, S, is either 1 or 2. Three two-digit numbers cannot add up to give a number 300 or more since 99 + 99 + 99 = 297. We have no information on what the value of R can be. All we know is that R cannot be 0 because 0+0+0 = 0 but V needs to be different from R.

Let’s look at the statements now.

Statement 1: M, N and P are positive even integers.

At first, it may seem that this has nothing to do with the value of R but we must analyze what is given to be sure.

M, N and P must take distinct values out of 2, 4, 6 and 8 and add up to give the units digit of T (again, distinct)

Every time you add three even numbers, you will get an even number. Let’s see which combinations we can get:

2 + 4 + 6 = 12

2 + 4 + 8 = 14

2 + 6 + 8 = 16

4 + 6 + 8 = 18

Note that in all four cases, the units digit is one of the numbers but T must be distinct. This means that there must have been a carryover from the previous addition. So when we added the three Rs, we must have got a carryover. Had R been 3 or less, we would not have got a carryover since 1+1+1 = 3, 2+2+2 = 6 and 3+3+3 = 9. So R must be greater than 3.

One such case would be

https://www.veritasprep.com/blog/wp-content/uploads/2015/04/Alphametic1.jpg

This statement alone is sufficient.

Statement 2: S = 2

The result of addition gives us a number which is more than 200. In statement 1 we saw a case in which S is 2 and R is greater than 3. Now all we have to do is find a case in which S is 2 and R is less than 3. One of these cases is

https://www.veritasprep.com/blog/wp-content/uploads/2015/04/Alphametic2.jpg

So this statement alone is not sufficient.

Answer (A)   This question is discussed   HERE

PART II 
 PART III 
 PRACTICE QUESTIONS

Bunuel · Accepted Answer

Alphametic Questions to Practice from Special Questions Directory:

PS:

DS:

https://gmatclub.com/forum/in-the-correc ... 28953.html
https://gmatclub.com/forum/if-in-the-add ... 97479.html
https://gmatclub.com/forum/in-the-correc ... 75564.html
https://gmatclub.com/forum/in-the-correc ... 97408.html
https://gmatclub.com/forum/in-the-additi ... 90569.html

Bunuel · Answer

AN INTRODUCTION TO SOLVING ALPHAMETIC QUESTIONS ON THE GMAT  
  BY KARISHMA, VERITAS PREP

Above, we looked at alphametics involving addition and subtraction. The logic becomes a little more involved when the alphametic involves multiplication. When a two digit number is multiplied by another two digit number, the process of finding the result is composed of multiple levels. Today, let’s see how to handle those multiple levels.  The question involves quite a few steps and observations using number properties. Hence, you are unlikely to see such a question in actual GMAT but you might see a simpler version so it’s good to be prepared.

Question:  The following alphametic shows multiplication of two numbers, IF and DR. The product you obtain is DORF.

https://www.veritasprep.com/blog/wp-content/uploads/2015/04/AlphameticPost2Fig1.jpg

What is the value of D + O + R + F?

(A) 17
(B) 20
(C) 22
(D) 23
(E) 30

Solution:   The good thing is that we know D + O + R + F has a single value. This means there will be a logic to obtain the value of each of D, O, R and F.

As discussed above, we first focus on the big picture, but we will have to go one level at a time.

(i) IF * R = OFF

(ii) IF * D = IF

(iii) OF + IF = DOR

A few interesting points to note from the above:

– From (ii), IF * D = IF

When you multiply IF by D, you get IF itself. This means that D must be 1. D can take no other value.

D = 1

– From (iii), F + F has unit’s digit of R.

Also O + I gives O as unit’s digit and 1 as tens digit (D of DORF obtained from above). How can this happen? Say, if O = 4, 4 + I = 14. This is possible only when I = 9 and there is a 1 carry over from the previous addition of F + F. This means that F must be 5 or greater to have a carryover of 1. It cannot be 5 because 5+5 will give you 10 making R = 0. This would mean that F*R would end in R (0). But in (i), F * R has unit’s digit of F, not R. So F cannot be 5.

D = 1, I  = 9

– Another interesting point: From (i), F * R has unit’s digit of F. This is possible only when F = 0 or F = 5 or R = 6 (Think of multiplication tables of numbers to convince yourself why this is so)

Since F has to be greater than 5 (as seen above), R must be 6.

If R = 6, then from (iii), F + F has unit’s digit of 6 and a carryover of 1 so F = 8. When you add 8 + 8, you will get 16 (units digit of 6 and a carryover)

D = 1, I = 9, R = 6, F = 8

– From (i), when we multiply IF by R, we get OFF. That is, when we multiply 98 by 6, we get 588. So O must be 5.

This gives us: D + O + R + F = 1 + 5 + 6 + 8 = 20

Answer (B)   This question is discussed   HERE .

This question uses your understanding of numbers and how they are added and multiplied. It certainly takes time to get to the answer. Such questions can help you get a feel for numbers and their behavior.

Bunuel · Answer

Is this GMAT Question an Alphametic or Simple Number Properties Question?  
  BY KARISHMA, VERITAS PREP

As noticed in the first post of Alphametics, a data sufficiency alphametic is far more complicated than a problem solving alphametic. An alphametic can have multiple solutions and establishing that it does not, is time consuming. Hence, it is less likely that you will see a DS alphametic in the actual exam.

In fact, what may look like an alphametic problem, might actually be a number properties problem only.

We will look at an example below:

Question:

https://www.veritasprep.com/blog/wp-content/uploads/2015/05/AlphameticPost3.jpg

In the correctly-worked multiplication problem above, each symbol represents a different nonzero digit. What is the value of C?

Statement 1: D is prime.
Statement 2: B is not prime.

Solution:   We multiply two two-digit integers and get 1995. The good thing is that we know the result of the multiplication will be 1995. Usually, multiplication alphametics are harder since they involve multiple levels, but here the multiplication is actually a blessing. There are many many ways in which you can ADD two integers to give 1995 but there are only a few ways in which you can multiply two integers to give you 1995.

Let’s prime factorize 1995:

1995 = 3*5*7*19

We can probably count on our fingers the number of ways in which we can select AB and CD.

19 needs to be multiplied with one other factor to give us a two digit number since 5*3*7 = 105 (a three digit number) so AB and CD cannot be 19 and 105.

19*3 = 57, 5*7 = 35 – This is not possible since two of the four digits are same here – 5.

19*5 = 95, 3*7 = 21 – This is one option for AB and CD.

19*7 = 133 – Three digit number not possible.

Hence AB and CD can only take values out of 21 and 95.

As of now, C can be 2 or 9. We need to find whether the given statements give us a unique value of C.

Statement 1:  D is prime

D is the units digit of CD. So D can be 1 or 5.

1 is not prime so CD cannot be 21. Hence, CD must be 95 and AB must be 21.

Hence, C must be 9.

This statement alone is sufficient.

Statement 2:  B is not prime

If B is not prime then AB cannot be 95. Hence AB must be 21.

This means CD will be 95 and C will be 9.

This statement alone is sufficient.

Answer (D)   This question is discussed   HERE .

Note that the entire question was just about number properties – prime factors, prime numbers etc. Actually it required no iterative steps and no hit and trial. Rest assured that if it is a GMAT question, it will be reasoning based and will not require painful calculations.

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ALPHAMETIC QUESTIONS ON THE GMAT

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ALPHAMETIC QUESTIONS ON THE GMAT

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615 to 715 in 15 Days (Ria's Strategies)

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My Rewards