Find the number puzzle.

@abrakadabra • Oct 25, 2024

Oct 25, 2024

1.7K

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.

Like 0 Replies 8

Replies

Welcome, guest

Join CrazyEngineers to reply, ask questions, and participate in conversations.

CrazyEngineers powered by Jatra Community Platform

kedarjk

@kedarjk-R7FdZW • Dec 3, 2013

2519
I wrote a simple python program to find it out. 😉
0
Ankita Katdare

@abrakadabra • Dec 3, 2013

kedarjk
2519
I wrote a simple python program to find it out. 😉
Please share the script, friend! 😁
0
kedarjk

@kedarjk-R7FdZW • 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.
0
Anoop Kumar

@anoop-kumar-GDGRCn • 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.
0
Alok mishra

@alok-mishra-E4tgN8 • 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 .
0
Anoop Kumar

@anoop-kumar-GDGRCn • 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.
0
smorgasbord

@smorgasbord-Aq7fYb • 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.
0
smorgasbord

@smorgasbord-Aq7fYb • 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. 😛
0