Question

(6) 3. The letters a and b are digits fie.. 0.1. 9). Find three pairs of digits (a, b) so that 5a3141104 by 3, but is not divisible by 9. Find three pairs of digits (a, b) so that 5a31411b2 is divisible a = 6,12,15 B = 21, 24,30 (6,21) & (12,24) X (15,30) X (6) 4. George mad enough money by selling candy bars at 15 cents each to buy several cans of pop at 48cents each. If he had no money left over, what is the fewest number candy bars he could have sold? 1.80115 2425 WO 130 iz cans) cans 1. 80 wont $0.48×3 = $1.44 0.48x4=$1.92 by a whole number of cans with no money J left (5) 5. Find the least common multiple of 42, 126, and 70 without listing.

Answer 1

of the no. divides 3, then sum of digits will divide 3. As it doesn't diride a so, sum of digits will not divide q. so sta+3+1+4 +1+1+bt2 = 17 tarb 3/17 tatb) but alottatb) so, we have to consider a, b in such a way that 17 tatb is a multiple of 3 but not a. of atb=1 , then 31(17 tatlo) & 9/87+a+b), so atb #1. Neamest multiple of 3 after 18 is 21 then 24 such that 9821 8 al 24 . .. let, 17 ta+6=21 a+b=4 . ? (a = 0,624), (0-1, 63), (Q=2, b=2),la-3, 6-1) (a=4,620) so, 4 pairs of (alb) are (0,4), (1,3), (2,27 å (3,1) so that 3)503141162 ay 5a3141162 · 7. Let he sold a candies & bought y cans. Then 15*= 48y = N (say) for a be lowest satisfying above entera ,so, Nalcm (15,48) Multiples of 48 : 48,96,144,192, 240, 298,- Matiples of is 15,30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225, 240,--... so, N224015 x 16 So, fewest no. of candy should be sold is 16. let na lim (42, 126,70) multiple of To will have a factor 10 on will have to ao (zero) in right of it, so, we must final nearest multiple of the largest no. (126) so that it has a zero in its unit place. • 126 nos last digit 6,3o, at least s should be multiplied to get a zero in its unit plake. How, 12665 = 630 Now, 630 = 90 15 4 430 = 9 Scanned with Sob clem (427126,70) = 630 =N As eving multiple of to will a