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.
Let the original number is 82 i.e 8 + 2 = 10 and when the digits are reversed i.e 28
The difference is 82 - 28 = 54 i.e, the number is decreased by 54.
The weight will be 250g plus (1.55 - 0.65)/0.10 units of 100g
250 + 900 = 1150
This is the maximum weight that can be sent at that price. But, weights exceeding
250 + 800 will also get charged this amount (that is what the ?part thereof? implies).
Hence a package weighing 1145 will be charged $1145
, which is always divisible by 6 and 12 both, since n(n+1) is always even.
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.
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.
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.
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.
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.
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