Two prime numbers A,B(A
If A and B are twin primes with B>23, then which of the following numbers would always divide A+B?
Any prime number greater than 3 will be in the form of 6x+1 or 6x−1.
Thus, both prime number are twins:
Let first be 6x−1
and 2nd be 6x+1
Thus it is always divisible by 12.
A and B are playing mathematical puzzles. A asks B "the whole numbers, greater than one, which can divide all the nine three digit numbers 111,222,333,444,555,666, 777,888 and 999?"
B immediately gave the desired answer. It was:
Each of the number can be written as a multiple of 111.
The factors of 111 are 3 and 37
Thus the desired answer is 3, 37 and 111
The remainder when the positive integer m is divided by 7 is x. The remainder when m is divided by 14 is x+7.
Which one of the following could m equal?
No answer description available for this question.
How many five digit multiples of 11 are there, if the five digits are 3, 4, 5, 6 and 7?
Repetition of digits is now allowed?
A number is divisible by 11, if the difference between the sum of digits at even places and odd places is either 0 or divisible by 11.
Numbers 5,3,6,4,7 is a multiple of 11
∵(5+6+7)−(3+4)=11, which is a multiple of 11
The number at odd places 5,6,7 can be arranged in 3! ways and 3,4 can be arranged in 2! ways.
Therefore, the total number of ways of such numbers =3!×2!=12
If we write down all the natural numbers from 259 to 492 side by side get a very large natural number 259260261....491492259260261....491492.
How many 8's will be used to write this large natural number?
58, there are 200 natural numbers so there will be 2×20=40 8's
From 459 to 492 we have 13 more 8's and so answer is 40+13= 53