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## May 29, 2009

### Number Lock Problems

There is a safe with a 5 digit Number lock. The 4th digit is 4
greater than second digit, while 3rd digit is 3 less than 2nd digit.
The 1st digit is thrice the last digit. There are 3 pairs whose
sum is 11. Find the number ?

#### 6 comments:

Ruth in NC said...

29256

Nicole said...

65292

the catman said...

I got 65292.

If you set the digits to abcde, where a thru e are integers between 0 and 9, you can immediately get (from the clues) that c can only be 1 or 2, because if it were any greater than 2 this would result in d > 9, out of our range... and thus d ε {8,9}. Also, b ε {4,5} so you have two possibilities there:

(1) c = 1, d = 8, and b = 4
(2) c = 2, d = 9, and b = 5

Doing the same thing with the other two digits (a, e) you get three possibilities:

(1) e = 1, a = 3
(2) e = 2, a = 6
(3) e = 3, a = 9

Using the clue that three pairs add to 11, the only combination from the first group with the second group that gives you these three pairs (whose sums are 11) would be (2) from the first group and (2) from the second group, i.e.

c = 2, d = 9, b = 5 and
a = 6, e = 2.

Putting these back into the combination results in the answer: 65292.

suvendu said...

its 65292.......gud question

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ranjit kumar said...

no u all r wrong...
its 9^5/2