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SOLVED PROBLEMS Simple problems: 1.What least number must be added to 3000 to obtain a number exactly divisible by 19? Solution: On dividing 3000 by 19 we get 17 as remainder Therefore number to be added = 19-17=2. 2.Find the unit's digit n the product 2467 153 * 34172? Solution: Unit's digit in the given product=Unit's digit in 7 153 * 172 Now 7 4 gives unit digit 1 7 152 gives unit digit 1 7 153 gives 1*7=7.Also 172 gives 1 Hence unit's digit in the product =7*1=7. 3.Find the total number of prime factors in 411 *7 5 *112 ? Solution: 411 7 5 112= (2*2) 11 *7 5 *112 = 222 *7 5 *112 Total number of prime factors=22+5+2=29 4.The least umber of five digits which is exactly divisible by 12,15 and 18 is? a.10010 b.10015 c.10020 d.10080 Solution: Least number of five digits is 10000 L.C.Mof 12,15,18 s 180. On dividing 10000 by 180,the remainder is 100. Therefore required number=10000+(180-100) =10080. Ans (d). 5.The least number which is perfect square and is divisible by each of the numbers 16,20 and 24 is? a.1600 b.3600 c.6400 d.14400 Solution: The least number divisible by 16,20,24 = L.C.M of 16,20,24=240 =2*2*2*2*3*5 To make it a perfect square it must be multiplied by 3*5. Therefore required number =240*3*5=3600. Ans (b). 6.A positive number which when added to 1000 gives a sum , which is greater than when it is multiplied by 1000. The positive integer is? a.1 b.3 c.5 d.7 Solution: 1000+N>1000N clearly N=1. 7.How many numbers between 11 and 90 are divisible by 7? Solution: The required numbers are 14,21,28,...........,84. This is an A.P with a=14,d=7. Let it contain n terms then T =84=a+(n-1)d =14+(n-1)7 =7+7n 7n=77 =>n=11. 8.Find the sum of all odd numbers up to 100? Solution: The given numbers are 1,3,5.........99. This is an A.P with a=1,d=2. Let it contain n terms 1+(n-1)2=99 =>n=50 Then required sum =n/2(first term +last term) =50/2(1+99)=2500. 9.How many terms are there in 2,4,6,8..........,1024? Solution: Clearly 2,4,6........1024 form a G.P with a=2,r=2 Let the number of terms be n then 2*2 n-1=1024 2n-1 =512=29 n-1=9 n=10. 10.2+22+23+24+25..........+28=? Solution: Given series is a G.P with a=2,r=2 and n=8. Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2. =2*255=510. 11.Find the number of zeros in 27!? Solution: Short cut method : number of zeros in 27!=27/5 + 27/25 =5+1=6zeros. BACK |