Maths question I've been stressing over-

Question

Choosing from the numbers: {1, 2, 5, 6, 7, 9}

find which number each letter stands for:

LAY

+ GAY

----------

EGGS

Answer 1

This one was FUN!!!

Looking at the equation and the #s available, by trial and error, G is likely an ODD number.

And with the digits in the HUNDRED place, the only way to get G to repeat in the addend the way it does, is with regrouping from the TENS place so that L + 1 = 10. The only way that works if is L = 9.

Once this happens, you can determine that the L + G in the HUNDREDS place will result in a 2-digit number starting with 1...so E = 1.

Now with 2, 5, 6, 7 & 9 remaining...check Y + Y

Y can't be 7 because the answer would be 14 and 4 is NOT a number is the set

Y can be 6 because 2 is a number in the remaining set.

Y can't be 5 because 0 is NOT a number in the remaining set.

Say Y = 6

Then Y + Y = 12...making S = 2

With regrouping (from Y + Y),the sum of the digits in the TENS place will be A + A + 1 = G

Check the remaining numbers;

7 + 7 + 1 = 15...5 is a number in the remaining set

9 + 9 + 1 = 18...8 is NOT a number in the remaining set

So now A = 7 and G = 5

Summary 1 = E, 2 = S, 5 = G, 6 = Y, A = 7, L = 9

Check

976

576

---------

1552

Hope this helps...

Answer 2

Hi Becky,

These kind of questions are tricky, because there isn't a direct path to the solution. Instead, you have to just kind of make observations and see where they lead you. But first, let's make sure we understand the question.

Choosing from the numbers: {1, 2, 5, 6, 7, 9}, find which number each letter stands for

So we have six numbers {1, 2, 5, 6, 7, 9} and six letters {A, E, G, L, S, Y}. Each of those letters stands for one of those numbers, so maybe A=2, E=6, and so on. The only way we can know what each letter is, is that they solve the equation:

LAY

+ GAY

-----

EGGS

Now that we understand the question, let's make some observations. The first thing I notice, is that LAY and GAY are 3-digit numbers, but the sum is a 4-digit number (EGGS). Hold on, if that is the case, what could E possibly be? Even if we put the two biggest numbers (9 and 7) as L and G, that would give us 900 + 700 (because they are in the hundreds place) equals 1600. E must be 1 then, because there's no way the thousands place could go up to 2 or higher!

Okay, so we know E=1, what else can we observe? Well, I notice LAY and GAY both end in Y, which means the ones place is just a number plus itself (Y+Y), and then the ones place of that sum is S. Using this we can figure out what Y and S should be. How? S can only be 2, 5, 6, 7, or 9, so if the ones place of Y+Y ends up being anything other than those, Y can't be that number. So let's try each of those numbers as Y, and get our result:

Y=9 Y=7 Y=6 Y=5 Y=2

9 7 6 5 2

+ 9 + 7 + 6 + 5 + 2

--- --- --- --- ---

18 S=8 14 S=4 12 S=2 10 S=0 4 S=4

The only value of Y that gives us a valid S is Y=6, which yields S=2. So now we know E=1, Y=6, and S=2. Let's see what that leaves us with:

1 E=1, Y=6, S=2

LA6

+ GA6 Remaining: 5, 7, 9

-----

1GG2

Hey, look at that! The tens place is also a sum of the same number, A+A, plus the 1 carry from the ones place. That means we can use the same logic to figure out A and G:

A=9 A=7 A=5

1 1 1

9 7 5

+ 9 + 7 + 5

--- --- ---

19 G=9 15 G=5 11 G=1

We can conclude that A=7 and G=5, because 1 is already used, and 9 can't be both A and G! Which leaves us with one letter and one number unused: L and 9, so L=9.

And there we have it! A=7, E=1, G=5, L=9, S=2, and Y=6. You can double check this solution by plugging all the numbers in and verifying that the sum makes sense. I'll leave that last part up to you :)

Answer 3

E = 1 because at most you can carry over 1 from the hundred place (L+G).

L = 9 because the only way for L+G=G is if L = 9 & there's a carryover 1 from the tens place (A+A).

S needs to be even, so either S = 2 or S = 6.

S can't be 6 because that means Y needs to be 8, which is not in the set.

Therefore S = 2.

Now we also know either Y =1 or Y = 6.

We know E = 1 so it must be that Y = 6.

The only numbers left for A and G are 5 and 7.

We know that Y+Y=12 so there's a 1 carry over to the tens place (A+A).

Therefore A = 7 and G = 5.