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5 years, 2 months ago
5 years, 1 month ago
A number is divisible by 7, 11 and 13 when 6 digit number xyzxyz , so x= 4, y= 7 and z= 9 as a result (y+z):x= 4 Ans.
More detail Why & How
Step 1 : The LCM of 7, 11 and 13 is 1001.
Step 2 : In 6-digit number 479xyz , minimum values are x = 0, y=0 and z=0 i.e. assume the number as 479000.
Step 3 : By dividing 479000 with 1001 , we get Quotient = 478 and Rremainder 522.
Step 4 : (479000 - 522 = 478478 ), here 478478 is exactly divisible by 1001 , or 7, 11 and 13. But starting 3 digits are not 479.
Step 5 : So quotient must be 478 + 1 = 479
Step 6 : Then required 6-digits number is the Dividend = Divisor x quotient = 1001 x 479 = 479479
Step 7 : As we get the number 479xyz = 479479, hence x= 4, y = 7 and z = 9
Step 8 : Finally (y+z):x = (7+9):4= 4 is the answer.
5 years, 2 months ago
First, Let's find the LCM of 7,11,13.
LCM=1001
assume x=y=z=0
479000/1000=478+522/1001
So we have a reminder od 522 when we divided by 1001
So, we get the nearest multiple of above this all we have to do is add the difference between 1001 and 522
1001-522=479
479000+479=479479 which we known as a multiple of 1001.
So, one possible solution is
xyz=479
However, we're not quite done. Any number that's divisible by 1001is divisible by 7,11,13. So if we added 1001 to this number. It will satisfy our requirements
479479+1001=480480. So the other possible solution is
xyz=480
These are the only two possibilities.
5 years, 2 months ago
Any number is divisible by 7, 11, 13 when it is written in the form of xyzxyz. Hence x =4, y=7, z=9.
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