Question

When 1! + 2! + 3! +..... + 125! Is divided by 7, what is the remainder?

Answer 1

||✪✪ QUESTION ✪✪||

When 1! + 2! + 3! +..... + 125! Is divided by 7, what is the remainder ? [ Excellent Question ]

|| ✰✰ ANSWER ✰✰ ||

First of All Lets Understand The problem ,

we have to Find The remainder when sum of Factorial upto 125 is divisible by 7 .

Factorial :- A factorial is a function that multiplies a number by every number below it.

So, we can say That :-

→ 1! = 1

→ 2! = 1 * 2 = 2

→ 3! = 1 * 2 * 3 = 6

→ 4! = 1 * 2 * 3 * 4 = 24

→ 5! = 1 * 2 * 3 * 4 * 5 = 120

→ 6! = 1 * 2 * 3 * 4 * 5 * 6 = 720

→ 7! = 1 * 2 * 3 * 4 * 5 * 6 * 7 = 720*7

Similarly Next,

As we can see Here now, Factorial From 7 Have 7 Their, So, we can say That, All Next Terms From 7 have 7 , and They all are divisible by 7.

So, We just have to Look Upto 6! Sum ..

So,

→ Sum upto 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873 .

So, we can say That now,

→ ( 1! + 2! + 3! +..... + 125! ) / 7

→ (873)/7

→ 124*7 + 5

→ Remainder 5.

Hence, when The sum of Factorials upto 125! ( Or we can say that, upto infinity Factorial ) when Divisible by 7 , we get our Remainder as 5.

Answer 2

QUESTION :-

When 1! + 2! + 3! +..... + 125! Is divided by 7, what is the remainder?

SOLUTION :-

➠ first we solve,

➠ 1! = 1

➠ 2! = 1 ×2 = 2

➠ 3! = 1 × 2 × 3 = 6

➠4! = 1 ×2 × 3 × 4 = 24

➠ 5! = 1 ×2 × 3 ×4 ×5 = 120

➠6! = 1 × 2 ×3 × 4 ×5 ×6 = 720

➠ 7! = 1 × 2 × 3 ×4 × 5 ×6 ×7 = 720 × 7.

NOW,

➠ .°. Sum upto 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873.

➠.°. ( 1! + 2! + 3! +..... + 125! ) / 7

➠.°. (873)/7

➠ .°. 124 × 7 + 5

➠ $\tt\red{\boxed{ .°. \:Remainder \:is\: 5.}}$