Find the number puzzle.

Ankita Katdare · 2013-12-03T17:41:31+00:00

There 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.

Find the number puzzle.

Ankita Katdare

Administrator

Updated: Oct 25, 2024

Views: 1.4K

There 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.

0

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kedarjk

Member • Dec 3, 2013

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

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Ankita Katdare

Administrator • Dec 3, 2013

kedarjk
2519
I wrote a simple python program to find it out. 😉
Please share the script, friend! 😁

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kedarjk

Member • Dec 3, 2013

Its no brainer...
-----------------------------------------
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)

-----------------------------------------

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.

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Anoop Kumar

Member • Dec 3, 2013

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.

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Alok mishra

Member • Dec 5, 2013

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

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Anoop Kumar

Member • Dec 5, 2013

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.

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smorgasbord

Member • Apr 23, 2014

Ankita Katdare
There 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.

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smorgasbord

Member • Apr 23, 2014

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. 😛

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