Question

4. Find the greatest number that divides 2838, 4053, and
4458 leaving exactly 3 as remainder in each case.​

Answer 1

Qᴜᴇsᴛɪᴏɴ :- Find the greatest number that divides 2838, 4053, and 4458 leaving exactly 3 as remainder in each case. ?

ᴄᴏɴᴄᴇᴘᴛ ᴜsᴇᴅ :-

when we have to Find the greatest no. that will divide x, y and z leaving the same remainder in each case.

Than :-

☛ Required no. = H.C.F of (x - y), (y - z) and (z - x).

Sᴏʟᴜᴛɪᴏɴ :-

→ x = 2838

→ y = 4053

→ z = 4458

So,

→ (y - x) = 4053 - 2838 = 1215

→ (z - y) = 4458 - 4053 = 405

→ (z - x) = 4458 - 2838 = 1620

Now,

☞ HCF of (405, 1215, & 1620) :-

⟼ 405 = 5 * 3⁴

⟼ 1215 = 5 * 3⁴ * 3

⟼ 1620 = 5 * 3⁴ * 2²

☞ HCF = 5 * 3⁴ = 405 . (Ans.)

Hence, The Required Greatest Number is 405.

Answer 2

QUESTION :-

•Find the greatest number that divides 2838, 4053, and 4458 leaving exactly 3 as remainder in each case.

SOLUTION :-

•let assume ,

=> R = 2838

=>A = 4053

=>J = 4458

•Now,

=> (R - A) = (4053 - 2838) = 1215.

=> (J - A) = (4458 - 4053) = 405.

=> (J - R) = (4458 - 2838) = 1620

•Now, find HCF of (405,1215,1620).

=> 405 = 5 × 3 × 3 × 3 × 3

=> 1215 = 5 × 3 × 3 × 3 × 3 × 3

=>1620 = 5 × 3 × 3 × 3 × 3 × 2 × 2

=> .°. HCF = 5 × 3 × 3 × 3 × 3 = 405.

Therefore, the greatest number is 405.