Noshad wrote:
Let N=123456789101112.....4344 be the 79-digit number that is formed by writing the integers from 1 to 44 in order, one after the other. What is the remainder when N is divided by 45?
A. 1
B 4
C. 9
D. 18
E. 44
A LOGIC, which can get your probability of scoring high..
We are looking for division by 45, so any multiple of 45 will have 0 or 5 as units digit, and hence the units digit of remainder will be 4-0=4 or x4-5=9..
Thus possible answers will be 4 , 9 and 44..
Now let us solve it..
so let us find the remainder when N is divided by 9 as 45=5*9..
Sum of digits of 1234...4344
(1+2+3..9)+(10+11+12+13+..18+19)+(20+..29)+(30+..39)+(40+41+..44)=(1+2+3..9)+(1*10+1+2+3+..9)+(2*10+1+2+3+..9)+(3*10+1+2+3+..9)+(4*5+1+2+3+4)
=45+(10+45)+(20+45)+(30+45)+(20+10)=45*4+90, so the number N is divisible by 9....
Any number that we get as N-9x will also be divisible by 9..
But for N-9x to be divisible by 45, it has to be divisible by 5 too, so the units digit will be 0 or 5...N has a 4 in units digit, so when subtract 9 from it, the units digit will be 5. Thus 12345....4335 will be divisible by 45 and the remainder will be 9.
C
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