Find the number puzzle.

2519
I wrote a simple python program to find it out. 😉

kedarjk2519
I wrote a simple python program to find it out. 😉

Please share the script, friend! 😁

Its no brainer...
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for i in range (1,3000):
if i%9==8:
if i%8==7:
if i%7==6:
if i%6==5:
if i%5==4:
if i%4==3:
if i%3==2:
if i%2==1:
print "This is the number: "+ str(i)

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I was curious to check if there is really such number, so wrote a simple script. There must be some other logic to find it out.

That's an bad programming, number of loops will be 2519.

This is class 8th problem which is calculated by following method, It's like following

LCM (reminders) *x - GCD(Numbers) until you find the solution. x is an Integer.

I don't remember that exact method.

ianoopThat's an bad programming, number of loops will be 2519.

This is class 8th problem which is calculated by following method, It's like following

LCM (reminders) *x - GCD(Numbers) until you find the solution. x is an Integer.

I don't remember that exact method.

its n*LCM(divisor)-(difference Between divisor and remainder) , n is any natural number .

Another example of this question is
find x such that
reminder is 2 when divided by 3,
3 when divided by 4,
1 when it's 5.

Ankita KatdareThere is a number less than 3000 that when divided by 2 leaves a remainder of 1, when divided by 3 leaves a remainder of 2, when divided by 4 leaves a remainder of 3, When divided by 5 leaves a remainder of 4, when divided by 6 leaves a remainder of 5, and so on up to nine.

What is that number?
Please provide step by step solution to this problem.

A number when
divided by 2 leaves a remainder of 1 = divided by 2 leaves a remainder of 1
divided by 3 leaves a remainder of 2 = divided by 3 leaves a remainder of 1
divided by 4 leaves a remainder of 3 = divided by 4 leaves a remainder of 1
divided by 5 leaves a remainder of 4 = divided by 5 leaves a remainder of 1
.
.
.
. = divided by 9 leaves a remainder of 1

Which means we need LCM of 2,3,4,5 ( 60 ), 7 ( 420 ), 8 (420*2=840), 9 ( 840*3 ) = 2520 ... and something.

if we need all the right hand side of the ='s to be set right we need to add one to 2520, but we need all the left hand side of the ='s written above to be true, so need to subtract one from it.
2519.

Hell! i made an egregious error:
A number when
divided by 2 leaves a remainder of 1 = divided by 2 leaves a remainder of -1
divided by 3 leaves a remainder of 2 = divided by 3 leaves a remainder of -1
divided by 4 leaves a remainder of 3 = divided by 4 leaves a remainder of -1
divided by 5 leaves a remainder of 4 = divided by 5 leaves a remainder of -1
.
.
.
. = divided by 9 leaves a remainder of -1

Which means we need LCM of 2,3,4,5 ( 60 ), 7 ( 420 ), 8 (420*2=840), 9 ( 840*3 ) = 2520, & -1 = 2519
sorry. 😛