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Krunaal
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20 Dec 2024, 11:09
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Let the 5 digits be a, b, c, d, and e
Myrna: a+b = d+e = c
Deon: (10*a+b)/c =k and (10*d+e)/c = n where k and n are integers
Kermit: a<b<c>d>e
All the digits are unique, we need to find d and e in abcde
Let's start checking options,
d>e, so checking 3 and 0, c will be 3 which is not a unique integer => not possible
checking 4 and 3, c will be 7 which won't satisfy Deons observation => not possible
checking 5 and 4, c will be 9, a and b can be 3 and 6 => 36954 => this satisfies all observations.
Answer 5, 4
Myrna: a+b = d+e = c
Deon: (10*a+b)/c =k and (10*d+e)/c = n where k and n are integers
Kermit: a<b<c>d>e
All the digits are unique, we need to find d and e in abcde
Let's start checking options,
d>e, so checking 3 and 0, c will be 3 which is not a unique integer => not possible
checking 4 and 3, c will be 7 which won't satisfy Deons observation => not possible
checking 5 and 4, c will be 9, a and b can be 3 and 6 => 36954 => this satisfies all observations.
Answer 5, 4
20 Dec 2024, 11:28
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Let n = XYZAB be the number.
given, X+Y = A+B = Z , Z is factor of XY,AB , also the order - X<Y<Z>A>B, all are distinct digits.
Z is a digit from 0 - 9. For each digit from 0-9, we do the following:
Z = 0,1,2,3,4,5,6,7,8,9
We look for possible X,Y,A,B.
we need at least two distinct ways of expressing the digit as a sum of two integers, and the distinct numbers created using those integers should be multiples of Z)
Z = X + Y
0 = 0+0 ( n becomes 0)
1 = 1+0 ( only one distinct way of expressing as sum)
2= 1+1 (only one distinct way of expressing as sum)
3 = 1+2,2+1(Identical numbers expressed as sum)
4 = 1+3,2+2,3+1(not all digits are distinct)
5 = 1+4,2+3,3+2,4+1(the numbers formed using the digits, none of them are factors of 5)
6 = 1+5,2+4,4+2,1+5(there is one distinct number(42) that is a multiple of 6)
7 = 1+6, 2+5,3+4,4+3,2+5,6+1(none of the numbers formed any two digits from above pairs is multiple of 7)
8 = 1+7,2+6,3+6,4+4,5+3,6+3,7+1(none of the numbers formed any two digits from above pairs is multiple of 8)
9 = 1+8,2+7,3+6,4+5,5+4,6+3,7+2,8+1
All the numbers created using any two digits is a multiple of 9. So Z has to be 9.
One can see that from the given options, 954 can be the last three digits of n.
Note the order has to be 3rd digit > 4th digit > 5th digit, so, fourth digit is 5 and fifth digit is 4.
given, X+Y = A+B = Z , Z is factor of XY,AB , also the order - X<Y<Z>A>B, all are distinct digits.
Z is a digit from 0 - 9. For each digit from 0-9, we do the following:
Z = 0,1,2,3,4,5,6,7,8,9
We look for possible X,Y,A,B.
we need at least two distinct ways of expressing the digit as a sum of two integers, and the distinct numbers created using those integers should be multiples of Z)
Z = X + Y
0 = 0+0 ( n becomes 0)
1 = 1+0 ( only one distinct way of expressing as sum)
2= 1+1 (only one distinct way of expressing as sum)
3 = 1+2,2+1(Identical numbers expressed as sum)
4 = 1+3,2+2,3+1(not all digits are distinct)
5 = 1+4,2+3,3+2,4+1(the numbers formed using the digits, none of them are factors of 5)
6 = 1+5,2+4,4+2,1+5(there is one distinct number(42) that is a multiple of 6)
7 = 1+6, 2+5,3+4,4+3,2+5,6+1(none of the numbers formed any two digits from above pairs is multiple of 7)
8 = 1+7,2+6,3+6,4+4,5+3,6+3,7+1(none of the numbers formed any two digits from above pairs is multiple of 8)
9 = 1+8,2+7,3+6,4+5,5+4,6+3,7+2,8+1
All the numbers created using any two digits is a multiple of 9. So Z has to be 9.
One can see that from the given options, 954 can be the last three digits of n.
Note the order has to be 3rd digit > 4th digit > 5th digit, so, fourth digit is 5 and fifth digit is 4.
20 Dec 2024, 12:28
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Quote:
let the 5 digit number be abcde
given a+b=c
d+e=c
Also, the possible combinations of double digit numbers from the options are, in a way where the sum results in a single digit number in a decreasing order=
30,40,50,60,70,80,43,53,54.
All numbers ending with zero arent possible as their digit sum results in the tens digit, which is not possible as the numbers need to be unique. 43 and 53 are prime, hence d and e are 5 and 4
IMO Answer 5 and 4 in that order
The number= 36954
