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Least number which when divided by 35,45,55 and leaves remainder 18,28,38; is?

Problem 1: Least number which when divided by 35,45,55 and leaves remainder 18,28,38; is?
Solution: i) In this case we will evaluate l.c.m.
               ii) Here the difference between every divisor and remainder is same i.e. 17.
                  Therefore, required number = l.c.m. of (35,45,55)-17 = (3465-17)= 3448.

Problem 2: Least number which when divided by 5,6,7,8 and leaves remainder 3, but when divided by 9, leaves no remainder?
Solution: l.c.m. of 5,6,7,8 = 840
                 Required number = 840 k + 3
                 Least value of k for which (840 k + 3) is divided by 9 is 2
Therefore, required number = 840*2 + 3
                                            = 1683

Problem 3: Greater number of 4 digits which is divisible by each one of 12,18,21 and 28 is?
Solution: l.c.m. of 12,18,21,28 = 254
               Therefore, required number must be divisible by 254.
               Greatest four digit number = 9999
               On dividing 9999 by 252, remainder = 171
               Therefore, 9999-171 = 9828.

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The telephone bill of a certain establishment is party fixed and partly varies as the number of calls consumed. When in a certain month 540 calls made the bill is Rs.1800. In another month 620 calls are consumed then the bill becomes Rs.2040. In another month 500 units are consumed due to more holidays. The bill for that month would be :

 The telephone bill of a certain establishment is party fixed and partly varies as the number of calls      consumed. When in a certain month 540 calls made the bill is Rs.1800. In another month 620 calls are      consumed then the bill becomes Rs.2040. In another month 500 units are consumed due to more      holidays. The bill for that month would be : a) Rs.1560           b) Rs.1680      c) 1840            d) Rs.1950  Expl : Let the fixed amount be Rs. X and the cost of each unit be Rs. Y.     Then, 540y + x = 1800 …. And 620y + x =  2040     On subtracting (i) from (ii), we get 80y = 240 -> y = 3     Putting y = 3 in (i) we get :     540 * 3 +  x = 1800 x = (1800-1620) = 180     . :  Fixed charges = Rs.180, Charge per unit = Rs.3.     Total charges for consuming 500 units = 180 +(500*3) = Rs.1680