Math Riddle
Jc
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Pierr10 Posted messages 13848 Registration date Status Moderator Last intervention -
Pierr10 Posted messages 13848 Registration date Status Moderator Last intervention -
Je suis 8495.
5 answers
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Hello,
9995
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Best regards,
Jordane -
Hello
sum 32
32 - 5 = 27
Now to find three numbers whose sum is 27
only solution 9
9995
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Very difficult to catch a black cat in a dark room.
Especially when it's not there... -
Hello,
it would be too easy to give the answer directly =D, start from the constraint of "5", it is imposed so you must subtract it from 32 from the beginning to get the "available" sum for the remaining numbers, then you just need to see how this remainder is divisible to meet the other constraints...
Best regards. -
Hello
I'm arriving after the battle. Nevertheless, I'm providing the detailed solution:
We start by noticing that a 3-digit number is not compatible with the condition "sum of the digits = 32"
(the largest 3-digit number would be 995, whose digit sum is 23)
The sought number is therefore of the form abb5 (it must be less than 10000).
So a + 2b + 5 = 32
and a + 2b = 27
2b is an even number; a must therefore be odd.
We proceed by elimination in search of possible values for a
The values 1, 3, 5, 7 are not suitable because they give a 2-digit number for b.
Thus, the remaining value is 9 for a
9 + 2b = 27
2b = 18 and b= 9
The sought number is therefore 9995
What is well understood is clearly expressed,
And the words to say it come easily.
(Boileau) -
Not even funny, you need to let it think a little =D