100^2 - 99^2 + 98^2 - 97^2 + 96^2 - 95^2 + ... + 2^2 - 1^2 = ?
Since you can notice that there are fifty pairs of n^2 - (n-1) ^2,
n^2 - (n-1)^2 = n + (n - 1)
Thus 100^2 - 99^2 + 98^2 - 97^2 + 96^2 - 95^2 + ... + 2^2 - 1^2 can also be written as
100 + 99 + 98+ ... + 2 + 1 = (100 x 101)/2 = 5050
A number can be multiplied by multiple of nine
i.e 9 18 27 36 45 ...
and the resulting number consist of only one digit.
Can you identify the number ?
12345679
12345679 × 9 = 111111111 (only 1s)
12345679 × 18 = 222222222 (only 2s)
12345679 × 27 = 333333333 (only 3s)
12345679 × 36 = 444444444 (only 4s)
12345679 × 45 = 555555555 (only 5s)
Find three whole, positive numbers that have the same answer when multiplied together as when added together.
1,2, & 3
1 x 2 x 3 = 6 and 1 + 2 + 3 = 6
I know there are two methods by using three time the same number with an plus(+) operator , you can make sum as 60.
One of them is 20+20+20.
whats the other way ?
The Other way is 55+5
Can you find the least possible number such that
If the number is divided by 3 , it gives the remainder of 1;
If the number is divided by 4 , it gives the remainder of 2
If the number is divided by 5 , it gives the remainder of 3;
If the number is divided by 6 , it gives the remainder of 4.
58.
Approach
LCM of the numbers 3,4,5 & 6 is 60.
So if i divide 60 by any of the four number , it gives remainder as 0.
Now to get reminder as desired, is should be 2 short of 60. (3-1,4-2,5-3,6-4 )
There is one four-digit whole number n, such that the last four digits of n2 are in fact the original number n
Looking at the last digit, the last digit must be either 0, 1, 5 or 6.
Then looking at the last two digits, the last two digits must be either 00, 01, 25 or 76.
Then looking at the last three digits, the last three digits must be either 000, 001, 625 or 376.
Then looking at the last four digits, the last four digits must be either 0000, 0001, 0625 or 9376.
Out of those, only 9376 is a 4 digit number
A 3 digit number is such that it's unit digit is equal to the product of the other two digits which are prime. Also, the difference between it's reverse and itself is 396.
What is the sum of the three digits?
The required number is 236 and the sum is 11.
It is given that the first two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, first two digits must be either 2 or 3 i.e.
22, 23, 32 or 33 which means that there are four possible numbers - 224, 236,326 and 339.
Now, it is also given that - the difference between it's reverse and itself is 396.
Only 236 satisfies this condition. Hence, the sum of the three digits is 11.
A four digit number (not beginning with 0) can be represented by ABCD. There is one number such that ABCD=A^B*C^D, where A^B means A raised to the B power. Can you find it?
The only number is 2592. 2^5 is 32 and 9^2 is 81. 32*81 is 2592.
I know a 5 digit number having a property that With a 1 after it, it is three times as large as it would be with a one before it.
Guess the number ?
42857
Let the number be x
10x +1 = 3(100,000 + x)
=> x = 42857.
Using Only Five 5's and any mathematical operator make sum as 37
((5!+5!)-55)/5
=>(120+120-55)/5
=>185/5
=>37
5!/(5+5)+5*5
=>120/10 + 25
=>12 + 25
=>37