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Scorpion4ik [409]
2 years ago
6

A positive five-digit integer is in the form ; where , and are each distinct digits. What is the greatest possible value of that

is divisible by eleven
Mathematics
1 answer:
stiv31 [10]2 years ago
3 0

The the biggest 5 digit number based on the computation will be 87,978.

<h3>How to compute the value?</h3>

The difference between the sum of the odd-numbered digits (1st, 3rd, 5th...) and the sum of the even-numbered digits (2nd, 4th...) is divisible by 11.

An example is that 34871903 is divisible by 11

3+8+1+0=12

4+7+9+3=23

23-12=11

Here, we want (b + b) - (a + c + a) to be divisible by 11.

2b - (2a + c) to be divisible by 11

ab,cba (using 7 and 8 and 9 since they biggest)

78 987 --> 2*8 - (2*7 + 9) = 16 - (14 + 9) = 16 - 23 = -7 NO

87 978 --> 2*7 - (2*8 + 9) = 14 - (16 + 9) = 14 - 25 = -11 YES

79 897 --> 2*9 - (2*7 + 8) = 18 - (14 + 8) = 18 - 22 = -4 NO

97 879 --> 2*7 - (2*9 + 8) = 14 - (18 + 8) = 14 - 26 = -12 NO

89 798 --> 2*9 - (2*8 + 7) = 18 - (16 + 7) = 18 - 23 = -5 NO

98 789 --> 2*8 - (2*9 + 7) = 16 - (18 + 7) = 16 - 25 = -9 NO

Therefore, the biggest 5 digit number is 87,978.

Learn more about computations on:

brainly.com/question/4658834

#SPJ1

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