Here by trial and error method, we can obseve that
404404 = 404 x 1001; 415415 = 415 x 1001, etc.
So, any number of this form is divisible by 1001.
When we divide 1234 by 8, remainder is 2
When we divide 1235 by 8, remainder is 3
When we divide 1237 by 8, remainder is 5
---> 2 x 3 x 5 = 30
As 30 will not be the remainder because it is greater than 8,
when 30 divided by 8, remainder = 6.
When is divided by 9, we have
, remainder = 7
, remainder = 4
, remainder = 1
, remainder = 7
, remainder = 4
, remainder = 1
So, we have cyclicity of 3 factors i.e 7,4,1.
Hence only 3 remainders are possible.
Two digit numbers: The two digits can be 2 and 9: Two possibilities 29 and 92.
Three-digit numbers: The three digits can be 1, 2 and 9 => 3! Or 6 possibilities.
We cannot have three digits as (3, 3, 2) as the digits have to be distinct.
We cannot have numbers with 4 digits or more without repeating the digits.
So, there are totally 8 numbers.
To solve a pair of simultaneous equations such as those given we can add or subtract them.
Adding we get 4x + 4y = 20
Therefore 2x + 2y = 10
The value of 4 in 475 means the place value of 4 in 475 and not its face value. Face value means the digit itself though it is at any place in the given number. But place value means the value of digit in its place in the given number.
Here the place value of 4 in 475 can be determined by as 4 is in 100's place in 475.
Hence, the place value of 4 in 475 is 4x100 = 400.
Given number is 987 = 3 x 7 x 47.
So, required number must be divisible by each one of 3, 7, 47.
None of the numbers in 553681 and 555181 are divisible by 3. While 556581 is not divisible by 7.
Correct answer is 555681.
Last digit for the power of 6 is 6 (always)
Power cycle of 7 is 7, 9, 3, 1.
Now 467/4 gives a remainder of 3
Then the last digit is = 3
Last digit is 6 + 3 = 9.
Screws weight = 90x100 = 9000gms
bolts weight = 100 x 150 = 15000gms
weight = screws + bolts => 24000 gms =>24 kg
Given entire box weight = 35.5kg
empty box = entire box weight - weight => 35.5kg - 24kg => 11.5kg
so the empty box is 11.5kg.
Sum of decimal places = 7.
Since the last digit to the extreme right will be zero (since 4 x 5 = 20), so there will be 6 significant digits to the right of the decimal point.
1 to 300 = 300 numbers
squares btwn 1 to 300 = 1,4,9,16,25,36,49.....289 i.e, total 17
cubes = 1,8,27,64,125,216 i.e, 5
now 1 and 64 are common so total numbers that should be removed ( 17+5-2= 20 )
total 20 term removed so 300th term would be 320
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.