Math/Logic Problem Help - Substituting?

Hmm... I like these types of problems. Let's see:

If E is 0, then the last digit of the answer would be 0.

If E is 1, then the last digit of the answer would be 5.

If E is 2, then the last digit of the answer would be 0 (from 10).

If E is 3, then the last digit of the answer would be 5 (from 15).

And on and on, so that makes A = 5, since it can't be equal to 0 lol obviously.

So that makes the answer 55555.

And also, E would have to be an odd number since an even E gives a 0 as the last digit. Except E can't be 5, since that's A, so E is 1,3,7 or 9.

If E is 1, then the sum of all Es would be a single digit and not carry over to the D column. This can't be a possibility because there are 4 Ds, and you can't end up with a 5 for that column with 4 of the

same single digit number. Let's try out the possibilities for D:

0x4 = 0

1x4 = 4

2x4 = 8

3x4 = 12

4x4 = 16

Can't use 5.

6x4 = 24

7x4 = 28

8x4 = 32

9x4 = 36

So since E can't be 1 or 5, it has to be 3, 7 or 9.

So here are the carry over digits for E:

If E is 3, then the digit that carries over would be 1 (from 15).

If E is 7, then the digit that carries over would be 3 (from 35).

If E is 9, then the digit that carries over would be 4 (from 45).

This is gonna be a bit confusing lol. Going back to the possibilities for D, and adding the carry over digit from E to the D column:

If E is 3, then the carry over digit 1 could be added to a D of 1 (to give you 5 for the D column), or a D of 6 (to give you a 25 for the D column).

If E is 7, then the carry over digit 3 could be added to a D of 3 (to give you 15 for the D column), or a D of 8 (to give you a 35 for the D column).

If E is 9, then the carry over digit 4 could not be added to any of the numbers because the result would not equal a multiple of 5.

So, if you understood all that haha, then that means:

E = 3, D = 1 or 6

OR

E = 7, D = 3 or 8

Ok so now, you basically continue this same process for the rest of the columns. First figure out the possibilities for adding the C column:

0x3 = 0

1x3 = 3

2x3 = 6

3x3 = 9

4x3 = 12

Can't use 5.

6x3 = 18

7x3 = 21

8x3 = 24

9x3 = 27

You may have noticed that I didn't use 5 but I used all the other numbers that we were saving for D and E. This is because A has already been permanently established as 5. But the numbers for D and E are

only possibilities; we limited the possibilities for D and E but we haven't actually found out the exact number for them.

So we already know the numbers of the C column plus the carry over digit from the D column have to result in a multiple of 5. Now looking back at our possibilities for E and D, we also need to jot down the

carry over digit from the D column to the E column. You can just scroll up to find that out, in the part where I was saying "If E is 3, then the carry over digit 1 could be added to a D...." So that means:

E = 3, D = 1 (with no carry over) or 6 (with a carry over of 2)

OR

E = 7, D = 3 (with a carry over of 1) or 8 (with a carry over of 3)

Alright, so E=3, D=1 does not work out because without a carry over, the sum of the C column does not equal a multiple of 5.

If E=3 and D=6 (and don't forget A=5), then the possibilties for C are 0,1,2,4,7,8 and 9. Now out of all these different Cs, and adding a 2 (which is the carry over digit for D=6), the only C remaining which

gives us a multiple of 5 is C=1. So with this possibility, E=3, D=6, C=1, B=? and A=5. We still have to try out the other possibilities though, just in case. And by the way, remember that a C of 1 has no

carry over digit to the B column.

So our next possibility is E=7, D=3. This carries over a 1 to the C column, and the only way to get a multiple of 5 for the C column with a carry over of 1 is if C=8. So with this possibility, E=7, D=3, C=8,

B=? and A=5. And remember, a C of 8 has a carry over of 2 to the B column.

And our last possibility is E=7, D=8. This carries over a 3 to the C column, and the only way to get a multiple of 5 for the C column with a carry over of 3 is if C=4. So with this possibility, E=7, D=8,

C=4, B=? and A=5. And remember, a C of 4 has a carry over of 1 to the B column.

So our final 3 possibilites up to this point are:

E=3, D=6, C=1, B=? and A=5; no carry over to the B column.

E=7, D=3, C=8, B=? and A=5; a carry over of 2 to the B column.

E=7, D=8, C=4, B=? and A=5; a carry over of 1 to the B column.

And finally, let's lay out the sum of the B column:

0x2 = 0

1x2 = 2

2x2 = 4

3x2 = 6

4x2 = 8

Can't use 5.

6x2 = 12

7x2 = 14

8x2 = 16

9x2 = 18

A key thing to note at this point is that the sum of the B column can not have any carry over digit to the A column, common sense. So that means B can't be any number greater than 4. Now looking at the carry

over digits from the C column to the B column, a carry over of 2 does not work because the sum of the B column with a B value of 4 or less plus the carry over d

First, enable's set some variables. we are going to enable 't' be the style of tulips that Abigail offered; this means that 't+4' is the style of tulips that Steven offered. additionally, we are going to enable 'r' be the style of roses that Abigail offered; this means that 'r+sixty 4' is the style of roses that Steve offered. Now that we've parts, enable's communicate approximately value. it is going to probable be much less demanding to artwork with pennies so we don't would desire to problem with regard to the decimal factor. So, utilising this, Abigail spent 90t + 201r on her flowers (ninety cents in step with tulip and 201 cents in step with rose). Steven spent ninety(t+4) + 201(r+sixty 4) on his flowers. next, the suggestions given says the two one among them spent 15108 cents on flowers, so....... 90t + 201r + ninety(t+4) + 201(r+sixty 4) = 15108 90t + 201r + 90t + 360 + 201r + 12864 = 15108 180t + 402r + 13224 = 15108 180t + 402r = 1884 (dividing by way of 6 to make the numbers smaller) 30t + 67r = 314 jointly because it style of feels that we don't have something to bypass on right here, we actually do if we use some uncomplicated experience. all of us understand which you would be able to not have a fractional section or a decimal component to a tulip or a rose; we additionally understand which you would be able to not have adverse roses or tulips; consequently, in trouble-free terms effective finished type would be appropriate for solutions. additionally, all of us understand that no rely what type we use for 't', multiplying it by way of 30 will yield a style that leads to 0. consequently we are able to focus on the 'r' value to get a 4 as a final digit. Multiples of sixty seven that lead to the digit 4 AND are under 314 supplies us in trouble-free terms one logical decision for 'r': 2. (the subsequent best value for 'r' that leads to a 4 would be 12 and (sixty seven*12) would be larger than 314, giving us an impossible adverse value for 't'. utilising this uncomplicated experience, and substituting 2 for 'r' we get: 30t + sixty seven*2 = 314 30t + 134 = 314 30t = a hundred and eighty t = 6 So after all that, Abigail had 6 tulips and 2 roses; Steven had 10 tulips and sixty six roses. a million day in the past

52487

52487 + 2487 + 487 + 87 + 7 = 55555

change all the a's to x

x=x, 2b=x, 3c=x, 4d=x 5e=x

a=1/1 b=1/2 c=1/3 d=1/4 e=1/5

werks 4 me, u?