LCM of 4, 6, 8 and 10 = 120
120) 1000 (8
960
------
40
The least number of four digits which is divisible by 4, 6, 8 and 10 => 1000 + 120 - 40 = 1080.
First line will cut all other 14, similarly second will cut 13, and so on
Total = 14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105.
Prime Numbers :: Numbers which are divisible by only 1 and itself are Prime Numbers.
It's answer will be 91.
Because 91 can be divisible by 7,13,91,1.
It is quite clear that prime number should be divisible only by itself and by 1.
Given
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 x 0 + 1 + 1
Using BODMAS Rule,
As multiplication precedes addition, 1 x 0 = 0,
Now, 10 + 0 + 1 + 1 = 10 + 2 = 12.
Hence, 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 x 0 + 1 + 1 = 12.
Let the two-digit number be 10a + b
a + b = 12 --- (1)
If a>b, a - b = 6
If b>a, b - a = 6
If a - b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 - a = 3
Number would be 93.
if b - a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 - b = 3.
Number would be 39.
There fore, Number would be 39 or 93.
Lowest 4-digit number is 1000.
LCM of 3, 4 and 5 is 60.
Dividing 1000 by 60, we get the remainder 40. Thus, the lowest 4-digit number that exactly divisible by 3, 4 and 5 is 1000 + (60 - 40) = 1020.
Now, add the remainder 2 that's required. Thus, the answer is 1022.
Let the smallest number be x.
Then larger number = (x + 1365)
x + 1365 = 6x + 15
= 5x = 1350
x = 270
Smaller number = 270.
Let the numbers be x and y. Then, x + y = 20 and x - y = 8.
x^2 - y^2 = (x + y) (x - y) = 20 x 8 = 160.
As per given
7th stop only one was left and in each preceding stop twice the nos... there were so 6 stops before 7th.
so
Let the five consecutive odd numbers be x-4, x-2, x, x+2, x+4
According to the question,
Difference between square of the average of first two odd number and the of the average last
two odd numbers is 396
i.e, x+3 and x-3
Hence, the smallest odd number is 33 - 4 = 29.
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