VBODMAS & Cyclicity

Applying VBODMAS, 3+3÷3×3-3

=3+1×3-3

=3+3-3

=6-3

=3

Applying VBODMAS rule, the given expression simplifies to

(2+1×2+2-1+1×2+2)/(2-2×1+2×2-2×1-2×2) = (2+2+2-1+2+2)/(2-2+4-2-4) = 9/(-2) = -4.5

Applying VBODMAS and keeping in mind that brackets are solved after bar, we proceed as under:

28 – [8 – {6 – (7 – 6 + 3)}]

= 28 – [8 – {6 – (7 – 9)}]

= 28 – [8 – {6 – (-2)}]

= 28 – [8 – {6 +2}]

= 28 – [8 – 8]

= 28

The last digit of 9823 is 3. Hence, the unit digit of (9823)⁴⁵ is decided by the unit digit of 3⁴⁵.

Now, 3¹ ends in 3, 3² ends in 9, 3³ ends in 7, 3⁴ ends in 1, 3⁵ ends in 3, 3⁶ ends in 9, 3⁷ ends in 7, 3⁸ ends in 1 etc.

The last digit repeats in sets of 4 and the pattern is: 3, 9, 7, 1

Now, 3⁴⁵ = 3⁴⁴*3

3⁴⁴ will end in 1, because the last digit of 3^x repeats in sets of 4 in the pattern 3,9,7,1.

-> Last digit of 3⁴⁵ = last digit of 3⁴⁴ * 3 = 1*3 =3

Therefore, (9823)⁴⁵ will also end in 3.

 

Alternately

The last digit of abc^x is the same as the last digit of c^x.

• Find out the cycle of last digits of c^xand make a list of those values.
• Find out the cyclicity, k
• Find out the remainder when the index x is divided by the cyclicity i.e. Rem[x/k] = r
• The answer would be the r^thvalue in the list; if r = 0, it would be the last value in the list.

Referring to the concept note above, we get, c = 3, x = 45, k = 4, and r = Rem[45/4]  = 1

The list is 3,9,7,1

-> The r^th value in this list is the 1^st value, which corresponds to 3.

Now, 2¹ ends in 2

2² ends in 4

2³ ends in 8

2⁴ ends in 6

2⁵ again ends in 2 etc.

Thus, the cyclicity of the last digit of 2⁷³⁷ is 4, and the pattern in which the last digit is repeated is: 2,4,8,6

Referring to the concept note below on finding the last digit of abc^x, we get, c = 2, x = 737, k = 4, and r = Rem [737/4] = 1

The list is 2,4,8,6

-> The r^th value in this list is the 1^st value, which corresponds to 2.

Note:

The last digit of abc^x is the same as the last digit of c^x.

• Find out the cycle of last digits of c^xand make a list of those values.
• Find out the cyclicity, k
• Find out the remainder when the index x is divided by the cyclicity i.e. Rem[x/k] = r
• The answer would be the r^thvalue in the list; if r = 0, it would be the last value in the list.

Last digit of 784569¹⁴² is the same as the last digit of 9¹⁴²

Now, 9¹ ends in 9

9² ends in 1

9³ again ends in 9

9⁴ again ends in 1

Thus, the cyclicity of the last digit of 9¹⁴² is 2, and the pattern in which the last digit is repeated is: 9,1

Referring to the concept note on finding the last digit of abc^x, we get, c = 9, x = 142, k = 2, and r = Rem [142/2] = 0

The list is 9,1 -> Since remainder is 0, the last digit corresponds to the second value in the list i.e., 1

 

Note:

The last digit of abc^x is the same as the last digit of c^x.

• Find out the cycle of last digits of c^xand make a list of those values.
• Find out the cyclicity, k
• Find out the remainder when the index x is divided by the cyclicity i.e. Rem[x/k] = r
• The answer would be the r^thvalue in the list; if r = 0, it would be the last value in the list.

Last digit of the given expression will be the same as the last digit of 4²⁴⁷*3⁶²⁹

Now, odd powers of 4 end in 4 -> 4²⁴⁷ will end in 4

Further, 3¹ ends in 3; 3² ends in 9; 3³ ends in 7; 3⁴ ends in 1; 3⁵ again ends in 3 etc. Thus, the cyclicity of the last digit of 3⁶²⁹ is 4, and the pattern in which the last digit is repeated is: 3,9,7,1

Referring to the concept note on finding the last digit of abc^x, we get, c = 3, x = 629, k = 4, and r = Rem [629/4] = 1

The list is 3,9,7,1

-> The r^th value in this list is the 1^st value, which corresponds to 3

-> last digit of 3⁶²⁹ is 3

Thus, last digit of 4²⁴⁷*3⁶²⁹ = last digit of (4*3) or last digit of 12 = 2

Note:

The last digit of abc^x is the same as the last digit of c^x.

• Find out the cycle of last digits of c^xand make a list of those values.
• Find out the cyclicity, k
• Find out the remainder when the index x is divided by the cyclicity i.e. Rem[x/k] = r
• The answer would be the r^th value in the list; if r = 0, it would be the last value in the list.

6³⁴³ -> ends in 6 (all powers of 6 end in 6)

9¹⁷² -> ends in 1 (even powers of 9 end in 1)

2¹²⁴ -> ends in 6 (applying concept in Q5)

7¹⁴⁴ -> ends in 1 (because pattern is 7,9,3,1 and cyclicity is 4)

Thus, last digit of 16³⁴³+19¹⁷²-32¹²⁴-27¹⁴⁴ = last digit of 6+1-6-1 = 0

Last digit of the given expression will be the same as the last digit of 8^{5^96}

Now, 5^{1 =} 5; 5² = 25; 5³ = 125; 5⁴ = 625; 5⁵ = 3125; 5⁶ = 15625 etc.

Thus, powers of 5x, where x is ≥ 3 end in 25 -> 5⁹⁶ ends in 25

Thus, 8^{5^96} can be written in the form 8^abc…..25

To find the last digit of 8^abc…..25, apply the cyclicity concept used in the above questions.

Here, the pattern is 8,4,2,6; cyclicity is 4, and the remainder r = Rem [abc…25/4] = 1 (remainder of a number divided by 4 is the same as the remainder of the number formed by the last two digits of that number; in this case number formed by the last two digits is 25

-> remainder is same as that obtained by dividing 25 by 4).

The list is 8,4,2,6

-> The r^th value in this list is the 1^st value, which corresponds to 8

-> last digit of 8^abc…..25 is 8

 

Note:

The last digit of abc^x is the same as the last digit of c^x.

• Find out the cycle of last digits of c^xand make a list of those values.
• Find out the cyclicity, k
• Find out the remainder when the index x is divided by the cyclicity i.e. Rem[x/k] = r
• The answer would be the r^th value in the list; if r = 0, it would be the last value in the list.

x^{y^z} ends in 1 (the only digit which is neither prime nor composite)

Now only 3 possibilities exist which satisfy the condition x:y:z = 1:2:3 -> 1^{2^3} , 2^{4^6}, 3^{6^9}.

However, 2^{4^6} ends in an even digit and is therefore ruled out. So we need to check for 1^{2^3} &  3^{6^9} for satisfying the given conditions. Now 1^{2^3} is 1 and therefore satisfies both the conditions, leaving only 3^{6^9} for verification.

We know that an even digit raised to power of a natural number > 1 will always be a multiple of 4.

Therefore,3^{6^9} = 3^4x

Cyclicity of 3^k where k is a natural number, is 4, and the pattern is 3, 9, 7, 1. So 3^4x ends in 1.

Therefore, 2 ordered triplets are possible which satisfy the given conditions -> (1, 2, 3) & (3, 6, 9)