# In the following sum the digits 0 to 9 have all been used, O = Odd, E = Even, zero is even and the top row's digits add to 9. Can you determine each digit?

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by saymruthravy

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by saymruthravy

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

423+675=1098 is the answer.

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

You have forgotten to give the puzzle. I have formulated the puzzle. I am giving the answer also.

E E O

E O O

===========

O E O E

===========

let the numbers be : a b c and d e f and the sum be : 1 g h i. we know the the thousands digit of the sum is 1. Other wise it will not be possible.

1. a+b+c = 9 => there are three combinations possible for a, b and c:

2,3,4 ;; 1, 3, 5 and 0, 2, 7

we hoped that digit 1 is in the sum and not the addends. so two combinations remain.

2. c ≠ 0 and f ≠ 0. as then i = c or f.

let us take combinations of 2,0,7.

so 720 , 270 not right. we cannot have d = 0, as then sum d+g will be 10 at most. Digit 0 is already used. That leaves us: 702, 207. with 207, the d has to be 8 or 9. which cases 1 and 2 will repeat in the sum. In case of 702, the middle digit needs a carry. So f = 8 or 9. In that case, i will be 0 or 1. This is ruled out.

Finally we are left with a b c = 2 3 4, 243, 324, 342, 423, 432.

d =2, gives rise to repetition of 0 or 1. Try the remaining combinations. we get the following.

One answer possible:

4 3 2

6 5 7

========

1 0 8 9

E E O

E O O

===========

O E O E

===========

let the numbers be : a b c and d e f and the sum be : 1 g h i. we know the the thousands digit of the sum is 1. Other wise it will not be possible.

1. a+b+c = 9 => there are three combinations possible for a, b and c:

2,3,4 ;; 1, 3, 5 and 0, 2, 7

we hoped that digit 1 is in the sum and not the addends. so two combinations remain.

2. c ≠ 0 and f ≠ 0. as then i = c or f.

let us take combinations of 2,0,7.

so 720 , 270 not right. we cannot have d = 0, as then sum d+g will be 10 at most. Digit 0 is already used. That leaves us: 702, 207. with 207, the d has to be 8 or 9. which cases 1 and 2 will repeat in the sum. In case of 702, the middle digit needs a carry. So f = 8 or 9. In that case, i will be 0 or 1. This is ruled out.

Finally we are left with a b c = 2 3 4, 243, 324, 342, 423, 432.

d =2, gives rise to repetition of 0 or 1. Try the remaining combinations. we get the following.

One answer possible:

4 3 2

6 5 7

========

1 0 8 9