### Brain Teasers

# Arrogant Arnold and Witty Will

Fun: (2.53)
Difficulty: (2.81)
Puzzle ID: #27632

Submitted By: Swordoffury1392 Corrected By: norcekri

Submitted By: Swordoffury1392 Corrected By: norcekri

Arrogant Arnold and Witty Will are brothers. One day, they were going from New Jersey to Florida. Bored, they decided to ask each other brain teasers. After a while, they got bored with this, so Arnold decided to place a bet. He would ask Will one teaser, which he thought was very hard. If Will got it right, Arnold would clean his room for the rest of the year. However, if Will got it WRONG, he would have to clean Arnold's room for a year and do his dishes for a month. After thinking for a while, Will agreed to this. This was Arnold's teaser:

Use each digit 1-9 only once each, so that one number multiplied by another equals the third. One example is 12*483=5796. Name at least one more!!!

Being witty as his name suggests, Will easily answered this. The question is, can you find one?

Use each digit 1-9 only once each, so that one number multiplied by another equals the third. One example is 12*483=5796. Name at least one more!!!

Being witty as his name suggests, Will easily answered this. The question is, can you find one?

### Hint

1) They can't be -12*483=-5796, 1.2*48.3=57.96, etc.2) The third number in another possibility is also 5796.

### Answer

The other possibilities are [drum roll please :)]...18 , 297 , 5346

27 , 198 , 5346

28 , 157 , 4396

39 , 186 , 7254

42 , 138 , 5796

48 , 159 , 7632

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## Comments

Good teaser! You got me! I am so not good at math teasers! Good work tho!

Thanks brainy, but could some more people give me comments please!

too much for my poor pea brain!

Not a bad teaser! I got it, but only after looking at part b of the clue, then I just tried dividing the answer by a 2 digit number that would allow for a 6 to be the last digit of the answer.

Is there a logical way of doing this besides trial and error?

I was able to reduce the problem before going to trial and error, but there were still a lot of possibilities.

The answer has to be in the form AB x CDE = FGHI. This means that A*C < 10. The possibilities for A and C are (1,{2..9}), (2,3), and (2,4).

(1,9) is easily eliminated since F would also have to be 9. With a little checking (1, and (2,4) can also be eliminated as they can't make two numbers low enough to not overflow into a 5-digit result. This leaves seven pairs of numbers for the first two digits times the two ways to assign these digits to A and C.

Neither B nor E can be 1 or 5 since that will make I either zero or B or E. This gives 7 * 6 = 42 combinations. If either B or E is a 6, then the other one has to be an odd number, so this eliminates (6,2), (6,4) and (6, times the two ways to assign the digits to B and E. Once B and E are assigned, then I is also determined. So there are 36 possible combinations for B, E and I.

Once A, B, C, E and I are assigned, there are four possibilities left for D, and assigning D determines F, G and H.

The above constraints limit the possible combinations to:

14 * 36 * 4 = 2016 possibilities.

This can be reduced quite a bit further because once A and C are choosen, all of the combinations of B, E and I that include either A or C are also eliminated. The following lists which of the 36 (B,E,I) combinations are eliminated for each possible assignment of (A,C):

(1,2): (3,7,1), (2,x,y), (3,4,2), (6,7,2), (8,9,2) - 22 eliminated

(1,3): (3,x,y), (7,9,3) - 16 eliminated

(1,4): (3,7,1), (4,x,y), (2,7,4), (3,8,4), (6,9,4) - 20 eliminated

(1,5): (3,7,1) - 1 eliminated

(1,6): 3,7,1), (6,x,y), (2,3,6), (2,8,6), (4,9,6), (7,8,6) - 13 eliminated

(1,7): (7,x,y) - 14 eliminated

(2,3): (2,x,y), (3,x,y), (4,8,2), (6,7,2), (7,9,3), (8,9,2) - 32 eliminated

Now the possibilites are down to:

2016 - (22 + 16 + 20 + 1 + 13 + 14 + 32)*4 = 1544

A little more checking finds that (1,7) for A and B can't make a small enough number to allow 9 to be used for either of the last two digits because F has to be 9, removing another (12 * 4) = 48 possibilites, leaving 1496.

That's still a lot of possibilites to search. I stopped when I found 18 x 297 = 5346.

The answer has to be in the form AB x CDE = FGHI. This means that A*C < 10. The possibilities for A and C are (1,{2..9}), (2,3), and (2,4).

(1,9) is easily eliminated since F would also have to be 9. With a little checking (1, and (2,4) can also be eliminated as they can't make two numbers low enough to not overflow into a 5-digit result. This leaves seven pairs of numbers for the first two digits times the two ways to assign these digits to A and C.

Neither B nor E can be 1 or 5 since that will make I either zero or B or E. This gives 7 * 6 = 42 combinations. If either B or E is a 6, then the other one has to be an odd number, so this eliminates (6,2), (6,4) and (6, times the two ways to assign the digits to B and E. Once B and E are assigned, then I is also determined. So there are 36 possible combinations for B, E and I.

Once A, B, C, E and I are assigned, there are four possibilities left for D, and assigning D determines F, G and H.

The above constraints limit the possible combinations to:

14 * 36 * 4 = 2016 possibilities.

This can be reduced quite a bit further because once A and C are choosen, all of the combinations of B, E and I that include either A or C are also eliminated. The following lists which of the 36 (B,E,I) combinations are eliminated for each possible assignment of (A,C):

(1,2): (3,7,1), (2,x,y), (3,4,2), (6,7,2), (8,9,2) - 22 eliminated

(1,3): (3,x,y), (7,9,3) - 16 eliminated

(1,4): (3,7,1), (4,x,y), (2,7,4), (3,8,4), (6,9,4) - 20 eliminated

(1,5): (3,7,1) - 1 eliminated

(1,6): 3,7,1), (6,x,y), (2,3,6), (2,8,6), (4,9,6), (7,8,6) - 13 eliminated

(1,7): (7,x,y) - 14 eliminated

(2,3): (2,x,y), (3,x,y), (4,8,2), (6,7,2), (7,9,3), (8,9,2) - 32 eliminated

Now the possibilites are down to:

2016 - (22 + 16 + 20 + 1 + 13 + 14 + 32)*4 = 1544

A little more checking finds that (1,7) for A and B can't make a small enough number to allow 9 to be used for either of the last two digits because F has to be 9, removing another (12 * 4) = 48 possibilites, leaving 1496.

That's still a lot of possibilites to search. I stopped when I found 18 x 297 = 5346.

In my previous comment, in the middle of the text reads as 8 )

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