Eight consecutive three-digit positive integers have the following property: Each of them is divisible by its last digit. What is the sum of the digits of the smallest of the eight integers ?
Answers
Given : Eight consecutive three-digit positive integers have the following property: Each of them is divisible by its last digit.
To find : sum of the digits of the smallest of the eight integers
Solution:
Possible numbers are
AB1 , AB2 , ................... , AB8
Divisible by 1 , 2 , ...........8
or
AB2 , AB3 , .............. , AB9
Divisible by 2 , 3 , ........., 9
so AB2 , AB3 , .............. , AB8
are Divisible by 2 , 3 , ........., 8
A + B + 3 is Divisible by 3
=> A + B is multiple of 3 ( 3 , 6 , 9 , 12 , 15 , 18)
AB4 Divisible by 4 if B4 Divisible by 4
=> B = 2 , 4 , 6 , 8 , 0
AB7 is divisible by 7
if 100A + 10B + 7 is Divisble by 7
if 100A + 10B is Divisble by 7
if 10A + B is Divisible by 7
Possibility (A , B)
(1 , 4) , ( 2 , 1) , ( 2, 8) , ( 3, 5) , ( 4 , 2) , ( 4 , 9) , ( 5 , 6) , (6 , 3) , (7 , 0) , (7 , 7) , (8 , 4) , (9 , 1 )
B = 2 , 4 , 6 , 8 , 0
Hence Removing cases where B ≠ 2 , 4 , 6 , 8 , 0
=> (1 , 4) , ( 2, 8) , ( 4 , 2) , ( 5 , 6) , (7 , 0) , (8 , 4)
A + B is multiple of 3 ( 3 , 6 , 9 , 12 , 15 , 18)
Hence Removing cases where A + B is not multiple of 3
=> ( 4 , 2) , (8 , 4)
421/1 = 421
422 /2 = 211
423/3 = 141
424 /4 = 106
425/5 = 85
426/6 = 71
427/7 = 61
428/8 = 53.5 ( not satisfied)
841/1 = 841
842/2 = 421
843/3 = 281
844/4 = 211
845/5 = 169
846/6 = 141
847/7 = 121
848/8 = 106
849/9 = 94.3
Hence 841 to 848 are 8 consecutive three-digit positive integers
Each of them is divisible by its last digit.
Smallest of Eight integer = 841
Sum of Digits = 8 + 4 + 1 = 13
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