[Shirsendu Roy]

a1,a1+a2,.....,a1+a2+...a100

if any one of these 100 terms divisible by 100 then we are through.

otherwise suppose that none of them is divisible by 100, then each leaves a remainder 1,2,...99. Since there are 100 sums and 99 remainder, by Pegion hole principle PP1, two of the sums leaves the same remainder after division by 100,

then let a1+a2+....+am=b100+r

a1+a2+....+an=c100+r

That gives m<n, a_{m+1}+a_{m+2}+....+a_{n}=100(c-b)

or, 100 divides a_{m+1}+a_{m+2}+....+a_{n}.

[Shirsendu Roy]

this number has least value compared to the other numbers given here.

[Shirsendu Roy]

97yz-97z-97=19xyz equation A

here 97-z-1 <yz

or, yz-z-1 <yz

or, 97(yz-z-1)<97yz

or,19xyz <97yz

or, 19x <97

or, x<97/19

or, x=1,2,3,4,5 is equation B

gcd(19,97)=1

or,at least one of x,y,z is multiple of 97 not=x

let y=97m

or, 97(97mz-z-1)=19x97mz

or,(97mz-z-1)=19xmz

or,(97-19x)mz=z+1  is equation C

here (97-19x)>=2 since x=1,2,3,4,5

(97-19x)mz>=2mz>=2z>z+1

and we have a contradiction with C

or, y cannot be multiple of 97

then let z=97m then A

becomes 97(y97m-97m-1)=19xy97m

or,97ym-97m-1=19xym

or,(97y-97-19xy)m=1 is equation D

since these are positive integer quantities, D implies that both m and 97y-97-19xy=1, m=1 or, z=97

or, 97y-19-19xy=1

or,97y-19xy=98

or,(97-19x)y=98

or,y=98/(97-19x)

where from A we get 97-19x=78,59,0,21and 2 for x=0,1,2,3,4,5

since y is positive integer, it follows that the first four values not taken, x=5,97-19x=2,y=49,z=97.

[Shirsendu Roy]

here (1.001)^1000<2^9<3^6<1000<(1.01)^1000

largest among numbers (1.01)^1000

[Shirsendu Roy]

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