IIT JEE , AIEEE Forum - Mathematics , Algebra

Sumitc

Let N be any five-digit number..say y₁y₂y₃y₄y₅ then what is the max value of N/(y₁+y₂+y₃+y₄+y₅)?

delhitushar

this is simple

the answer is 10000.

for n\y1+y2+y3+y4+y5 to be max y1+y2+y3+y4+y5 should be minimum

so one y is 1 &others are 0 . so the max balue is

10000/1+0+0+0+0

=10000

mohit1

Tushar have you ever went to warangal?

edison

Are teh five digits y1, y2, y3, y4, and y5 all different or same or whatever?

malay

the answer is 10000.
suppose the given ratio is >10000
N>10000(y1+y2+y3+y4+y5)
which is impossible........

Sumitc

the numbers y1,y2,y3,y4,y5 are all different.

edison

To solve this problem here we follow the simple logic

let y1 + y2 + y3 + y4 + y5 = X

then, if X > 9

N/(y1 + y2 + y3 + y4 + y5) = N/X < 10000

But for X = 9, then to have the ratio N/X maximum N should be maximum which is possible only when N = 90000

so N/X = 10000

similarly when X = 8, 7, 6, 5, 3, 2 or 1

then N/X = 10000 is the maximum value.

Thus in no case N/X exceeds 10000

so maximum value of N/(y1 + y2 + y3 + y4 + y5) = 10000

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