Question

let us find the number divisible by 13 such that when that number is divided by 8,12,16 and20,it leaves 1 as remainder in all the case​

Answer 1

✪||Solution||✪

LCM of 8,12,16 and 20 = 80

(multiple of LCM + 1) ÷ 13 = number

Let, put the second multiple of 80 in

place of multiple of LCM

(160 + 1) ÷ 13

= 161 ÷ 13

161 is divisible by 13 not perfectly.

Let, put the third multiple of 80 in place of multiple of LCM

(240 + 1) ÷ 13

= 242 ÷ 13

242 is divisible by 13 not perfectly.

Let, put the fourth multiple of 80 in place of multiple of LCM

(320 + 1) ÷ 13

= 321 ÷ 13

321 is divisible by 13 not perfectly.

Let, put the fifth multiple of 80 in place of multiple of LCM

(400 + 1) ÷ 13

= 401 ÷ 13

401 is divisible by 13 not perfectly.

Let, put the sixth multiple of 80 in place of multiple of LCM

(480 + 1) ÷ 13

= 481 ÷ 13

481 is divisible by 13 perfectly. So, 481 is answer.

481 is answer

${\green{\boxed{\boxed{\bf{\bigstar\:\: Answer : 481\:\:\bigstar}}}}}$

Answer 2

Answer:

Answer is 481

Step-by-step explanation:

Step 1 : Find LCM of 8 , 12, 16, and 20

Step 2: Then accordingly proceed using property Dividend = (Divisor * Quotient ) + Remainder

#wbbse

For mathematics science and more follow @tutoras95

on Instagram

contents by

@amirsalim95