Question

❤ Hola Brainliacs ❤


Find the smallest positive integer n such that -


$\sf{ 1000 \leqslant n \leqslant 1100 }$


and


$\sf{ 1111^n + 1222^n + 1333^n + 1444^n }$


Is divisible by 10.


Best Of Luck Solving
​

Answer 1

We should be considering the following statements.

(1)  Possible number which can be the unit digit of the powers of the numbers ending in 1 is 1 only. Means all powers of numbers ending in 1 also end in 1. So no. of possible numbers is 1.

(2)  Possible numbers which can be the unit digit of the powers of the numbers ending in 2 are 2, 4, 8 and 6 in the order. No. of possible numbers is 4.

(3)  Possible numbers which can be the unit digit of the powers of the numbers ending in 3 are 3, 9, 7 and 1 in the order. No. of possible numbers is 4.

(4)  Possible numbers which can be the unit digit of the powers of the numbers ending in 4 are 4 and 6 in the order. No. of possible numbers is 2.

[Note:- The term "order" means that successive powers of a number end in the mentioned digit in the order taken.]

First we've to choose one number from each case above, of which their sum should be a multiple of 10.

We find three such possibilities. We analyse each possibility.

$\sf{1.\quad 1+2+3+4=10}$

This is due to the supposition that,

• $\sf{1222^n\equiv2\pmod{10}}$

• $\sf{1333^n\equiv3\pmod{10}}$

These imply $\sf{n\equiv1\pmod{4}}$ because of the statements (2) and (3) above.

• $\sf{1444^n\equiv4\pmod{10}}$

This implies $\sf{n\equiv1\pmod{2}}$ because of the statement (4).

By the two conditions, let,

$\longrightarrow\sf{n=4p+1=2q+1}$

$\longrightarrow\sf{n-1=4p=2q}$

Since LCM of 4 and 2 is 4,

$\Longrightarrow\sf{4\mid n-1}$

Least possible value of $\sf{n-1}$ for $\sf{1000\leq n\leq 1100}$ is,

$\longrightarrow\sf{n-1=1000}$

$\longrightarrow\underline{\sf{n=1001}}$

$\sf{2.\quad 1+4+9+6=20}$

This is due to the supposition that,

• $\sf{1222^n\equiv4\pmod{10}}$

• $\sf{1333^n\equiv9\pmod{10}}$

These imply $\sf{n\equiv2\pmod{4}}$ because of the statements (2) and (3) above.

• $\sf{1444^n\equiv6\pmod{10}}$

This implies $\sf{n\equiv2\pmod{2}}$ because of the statement (4).

By the two conditions, let,

$\longrightarrow\sf{n=4p+2=2q+2}$

$\longrightarrow\sf{n-2=4p=2q}$

Since LCM of 4 and 2 is 4,

$\Longrightarrow\sf{4\mid n-2}$

Least possible value of $\sf{n-1}$ for $\sf{1000\leq n\leq 1100}$ is,

$\longrightarrow\sf{n-2=1000}$

$\longrightarrow\underline{\sf{n=1002}}$

$\sf{3.\quad 1+8+7+4=20}$

This is due to the supposition that,

• $\sf{1222^n\equiv8\pmod{10}}$

• $\sf{1333^n\equiv7\pmod{10}}$

These imply $\sf{n\equiv3\pmod{4}}$ because of the statements (2) and (3) above.

• $\sf{1444^n\equiv4\pmod{10}}$

This implies $\sf{n\equiv1\equiv3\pmod{2}}$ because of the statement (4).

By the two conditions, let,

$\longrightarrow\sf{n=4p+3=2q+3}$

$\longrightarrow\sf{n-3=4p=2q}$

Since LCM of 4 and 2 is 4,

$\Longrightarrow\sf{4\mid n-3}$

Least possible value of $\sf{n-1}$ for $\sf{1000\leq n\leq 1100}$ is,

$\longrightarrow\sf{n-3=1000}$

$\longrightarrow\underline{\sf{n=1003}}$

The smallest positive integer among the three possibilities is $\sf{1001.}$

$\longrightarrow\underline{\underline{\sf{n=1001}}}$

Hence 1001 is the answer.

Answer 2

For 1,

all the powers increased gradually can give 1 possible number, i.e. 1.

(∴N ≡ 1)

• 1¹ = 1
• 1² = 1
• 1³ = 1
• 1⁴ = 1

For 2,

all the powers increased gradually can give 4 possible numbers (for units place), i.e. 2, 4, 8, & 6.

(∴N≡4)

• 2¹ = 2
• 2² = 4
• 2³ = 8
• 2⁴ = 16 (6 is considered for being in the unit's place)

For 3,

all the powers increased gradually can give 4 possible numbers (for units place), i.e. 3, 9, 7, & 1.

(∴N ≡ 4)

• 3¹ = 3
• 3² = 9

• 3³ = 27 (7 is considered for being in the unit's place)

• 3⁴ = 81 (1 is considered for being in the unit's place)

For 4,

all the powers increased gradually can give 2 possible numbers (for units place), i.e. 4, & 6.

(∴N ≡ 2)

• 4¹ = 4

• 4² = 16 (6 is considered for being in the unit's place)

• 4³ = 64 (4 is considered for being in the unit's place)

• 4⁴ = 256 (6 is considered for being in the unit's place)

And after 4, the sequence is getting repeated.

$\rule{190}2$

Arranging all the sequences again :-

→ 1  1  1  1

→ 2  4  8  6

→ 3  9  7  1

→ 4  6  4  6

Adding all the digits in order :-

1 + 2 + 3 + 4 = 10

The sum ends with 0.

1 + 4 + 9 + 6 = 20

The sum ends with 0.

1 + 8 + 7 + 4 = 20

The sum ends with 0.

1 + 6 + 1 + 6 = 14

The sum does not end with 0.

$\rule{190}2$

Now,

1000 is (4th in sequence) as 1000 = 4 × 250

Hence,

1001 , 1002 , 1003 all are divisible by 10.

The least integer out of 1001, 1002, & 1003 is 1001.

Hence, n = 1001.