IIT JEE , AIEEE Forum - Mathematics , Algebra

sangeetha5491

how many eight digit numbers formed from 1,3,4,5,6,7,8,9 will be multiples of 275?

akhil_o

1,3,4,6,8,9,7,5

now no has to be divisible by
275=25*11
so it has to end in either 25,50 , 75 or 00
hence last digits from above nos is 75

also it is divisible by 11
so sum of alternate digits must differ by multiples of 11

the nos are 1,3,4,6,8,9
sum of these is 31
also now since 7,5 are fixed
the difference between sums of groups
has to be
2,9 or 20
but diff of 2 is not possible
because there are no integral pairs such that
a+b=31, a-b=2(no even nos)
and also 20 diff i not possible...max difference=15

so difference is 9

a+b=31
a-b=9

a=20, b=9
now such a pair is only

9,8,3 and 1,4,6
9+8+3=20
1+4+6=11

so (9,8,3),7 and (1,4,6),5
this is of form

19486375, where 1,4,6 interchange in 3!=6 ways
and 2,8,3 interchange in 3! ways=6

So total numbers
in 3!*3! ways
=36 ways

hence 36 nos are possible

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