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Quote:

If a, b, c, d and e are five consecutive positive integers. What is the unit's digit of a^b∗c^d∗e?


(1) c = 5
(2) a + b is six less than e + d.

Ans (A)
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If a, b, c, d and e are five consecutive positive integers. What is the unit's digit of \(a^b∗c^d∗e\)?
Let the numbers are
a , a + 1(b), a + 2(c), a + 3(d), a + 4(e). \(a^b∗c^d∗e\) = \(a^{a+1}∗(a+2)^{a+3}∗(a+4)\)
So, basically a = ?

(1) c = 5
Since a + 2 = 5
a = 3.

SUFFICIENT.

(2) a + b is six less than e + d.
a + b = e + d - 6

Here it's clear that under given condition any value of a, b, e and d would satisfy the equation.

INSUFFICIENT.

Answer A.
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The question says that a,b,c,d,e are five consecutive integers so minimum value of a = 1.

Now take a = n, b = n+1, c=n+2, d=n+3, e =n+4

S1: c = 5.
n+2 = 5, n = 3
Sufficient.

S2: a+b = d+e - 6
n+n+1 = n+3 + n+4 -6
2n + 1 = 2n +1
Multiple values possible.
Insufficient

IMO A
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Given that a,b,c,d, and e are five consecutive positive integers, we are to determine the unit digits of a^b * c^d * e.

It is worth noting that the fact that question says a,b,c,d, and e are five consecutive positive integers does not mean that a is the first integer, b second integer, c third, d fourth, and e the fifth integer.

Statement 1: c=5.
Statement 1 says 5 is one of the five consecutive integers. This means the consecutive integers can start from 5, end on 5, etc. In other words, 5 can be any of the numbers as long as it is part of the numbers.
If the five numbers are a=3,b=4,c=5,d=6, and e=7, such that the order is a,b,c,d,e.
then we have 3^4 * 5^6 * 7 = 1*5*7=35, hence the unit digit is 5.
If a=3,b=4,c=5,e=6, and d=7, implying the order is a,b,c,e,d,
then we have 3^4 * 5^7 * 6 = 1*5*6 = 30, hence the unit digit is 0.
Since we have two different unit digits, statement 1 is not sufficient.

Statement 2: a + b is six less than e + d.
This statement implies (e+d)-(a+b)=6.
Let a=3,b=4,c=5,d=6, and e=7 implying the order is a,b,c,d, and e.
a+b=7, and d+e=13, and 13-7=6 hence condition in statement 2 is satisfied.
3^4 * 5^6 * 7 = 1*5*7 = 35 hence unit digit is 5.
Let a=3,b=4,c=5,e=6, and d=7, implying the order is a,b,c,e,d. Then a+b=7, and e+d=13, and 13-7=6 hence statement 2 is statisfied.
3^4 * 5^7 * 6 = 1*5*6=30 with a unit digit of 0.
Statement 2 is not sufficient.

1+2.
Combining both statements will still not be sufficient since no new information is provided. We can still rearrange the numbers in a manner that yields two different unit digits.

The answer is E in my view.
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If a, b, c, d and e are five consecutive positive integers. What is the unit's digit of \(a^{b}∗c^{d}∗e\) ?

(Statement1): c = 5
--> no matter whether they are in increasing order, the bases of exponents are odd numbers (--> \(3^{4}*5^{6}*7= 81*7*5^{6}\))
the units digit of \(5^{6}\) is ...5. --> Multiplying \(5^{6}\) by Any odd number cannot change the units digits of that number

--> The units digit of \(81*7*5^{6}\) is ...5
Sufficient

(Statement2) a + b is six less than e + d.
a+b = e+d -6
a+a+1= a+3+a+4 -6
--> 2a+1 = 2a +1

--> 5 consecutive positive integers begin with both any odd numbers and any even numbers

if they begin with even numbers, then \(2^{3}* 4^{5}*6 = 8*6*1024\)

--> the units digit of \(2^{3}* 4^{5}*6\) is EVEN number.

if they begin with odd numbers, it is already solved above ( the units digit of ... is ODD number)
Insufficient

The answer is A.
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If a, b, c, d and e are five consecutive positive integers. What is the unit's digit of ab∗cd∗eab∗cd∗e?


(1) c = 5
(2) a + b is six less than e + d.

Statement 1: C= 5
if ant even number *5 = 0 in units digit.

Statement 1 is sufficient

Statement 2

a+b+6= D+e

This could be any combination of numbers

3+4+6=6+7

or
13+14+6=16+17

No matter the combination the middle number will be either 5 or multiple of 5 hence the units digit will be 0

IMO D
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