LCM of 32, 36, 48 and 54 is 864
∴ the greatest number that should be subtracted from 10000 is 864
Hence, 10000 - 864 = 9136

Shortcut:
To find the least number which when divided by x,y and z leaves the remainders a,b and c respectively.
It is always observed that, (x-a)= (y - b) = (z-c) = Z(say)
∴ the required number = (LCM of x,y and z)- Z
x = 52
y = 78
z = 117
a = 33
b = 59
c = 98
Using these values in shortcut, we get (x-a)= (y - b) = (z-c) = Z(say)
(52 - 33) = (78 - 59 ) = (117 - 98 ) = 19
LCM of 52, 78 and 117 is 468
∴ the required number = 468 - 19 = 449

The LCM of 6, 7, 8, 9 and 10 = 2520
The greatest number of six digits = 999999
Dividing 999999 by 2520, we get 2079 as remainder
∴ the number divisible by 2520,
= 999999 - 2079
= 997920

LCM 2, 3, 4, 5 and 6 = 60
Moreover the difference between each divisor and the corresponding remainder is the same, which is 1.
∴ required number is of the form (60Z-1), which is divisible by 7 for the least value of Z.
Now, on dividing (60Z - 1) by 7, we get (4Z - 1) as the remainder.
We find the least positive number Z for which (4Z - 1) is divisible by 7. By inspection Z = 30 ∴ the required number = 4 x 30 - 1 = 119

LCM of 5, 7, 12 and 15 = 420
The greatest number of 6 digits = 999999
We can break this number into multiple of 420 as 420 x 2380 + 399
Hence, the greatest number of six digits that is exactly divisible by the aboe number is 420 x 2380 = 999600
Now, (5 - 3) = (7 - 5) = (12 - 10) = 2
Hence, subtracting 2 from this greatest number we shall get the required number which is therfore equal to 999598

Shortcut:
To find the least number which, when divided by x, y,and z leaves the same remainder R in each case.
Required number = (LCM of x,y and z) = R
Using the shortcut, we get
Required least number = (LCM of 2,3,4,5 and 6) + 1 = 60 + 1 = 61

LCM of 34, 38, 85 and 95 = 3230
Now, divide 7894135 by 3230, we obtain 15 as remainder and 2444 as the quotient.
But, according to the question the remainder should be 11.
Hence, the required smallest number that must be subtracted is 15 - 11 = 4

LCM of first twelve integers excepting unity is 27720
The required number is of the form (27720Z + 1) which leaves remainder 1 in each case.
Now, on dividing (27720Z + 1) by 17, we get (10Z + 1) as the remainder.
We find the lest positive number Z for which (10Z+1) us divisible by 17
By inspection Z = 5
Hence, the required number = 27720 x 5 + 1 = 138601

LCM of 4, 6, 10 and 15 = 60
The least number of six digits is 100000
Dividing 100000 by 60, we get 40 as remainder.
Also (60 - 40) = 20, the least number of six digits exactly divisible by each of the above numbers = 100000 + 20 = 100020
∴ the least number of six digits which will leave a remainder 2 when divided by each of the given numbers = 100020 + 2 = 100022

Shortcut: To find the greatest number that will divide x, y and z leaving the same remainder 'R' in each case.
Required number = HCF of (x - R), (y - R) and (z - R)
x = 410, y = 751, z = 1030 and R = 7
Using these values in shortcut, we get
Required number = HCF of (410 - 7), ( 751 - 7) and (1030 - 7) = 31