Question

the prime no. dividing 109 and leaving a remainder of 4 in each case are​

Answer 1

There are many numbers that divide 109 with a remainder of 4. List all two-digit numbers that have that property.

If there is a remainder of 4, then 109-4= 105 should be completely divisible by the two-digit numbers.

=> All two-digit divisors of 105.

Factors of 105 are 3,5,7.

Two digit divisors are then 3*5=15, 3*7=21, 5*7=35

The two-digit numbers that divide 109 with a remainder of 4 is 15,21 and 35.

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Answer 2

if there is a reminder of 4 , then 109 - 4 = 105 should be completely divisible by two digit number s

All 2 digit divisers of 105

factors of 105 are 3,5,7

2 digit divisors r then 3*5=15, 3*7 =21, 5*7 =35

the 2 digit numbers that divide 109 with reminder of 4 is 15 , 21 and 35