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ka+ 3=0910. Howmany toodigit numbers are divisible de 3. |
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Answer» To find:No of 2 digit number divisible by 3 A.P : 12,15........99 a=12,d=3,an=99,n=? an=a+(n-1)d 99=12+(n-1)3 99=12+3n-3 99=3n+9 9n+3=99 3n=90 n=90/3 n=30 Thus 30 2 digit number are divisible by 3 Numbersdivisible by 3are3, 6, 9, 12, …………….. Lowesttwo digitnumberdivisible by 3is 12 Highesttwo digitnumberdivisible by 3We know that 99/3= 33 ∴ Highesttwo digitnumberdivisible by 3is 99 So the series starts with 12 and ends with 99. Difference between numbers is3. AP., 12 , 15, , ,99; a=12, d=3; an=99, an=a+( n-1)d; 99=12+ ( n-1)3; 99-12=( n-1)3; 87=( n-1)3; 87/3= ( n-1); 29=( n-1); n=29+1=30 30 no. is divisible by 3 there are 33 2digit numbers are divisibleby 3 |
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