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Bunuel
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What is the last two digits of (201*202*203*204*246*247*248*249)^2 ? 19 Sep 2019, 03:40
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

38% (02:06) correct 62% (01:56) wrong based on 119 sessions

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What is the last two digits of \((201*202*203*204*246*247*248*249)^2\) ?

A. 10
B. 16
C. 36
D. 56
E. 76


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Find the last 2 digits
(201*202*203*204*246*247*248*249)^2
1.36
2.56
3.76
4.16
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E
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What is the last two digits of (201*202*203*204*246*247*248*249)^2 ? 19 Sep 2019, 04:17
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Last 2 digits of \(201^2\)= \(01^2=01\)
Last 2 digits of \(202^2\)= \(02^2=04\)
Last 2 digits of \(203^2\)=\(03^2=09\)
Last 2 digits of \(204^2\)= \(04^2=16\)
Last 2 digits of \(246^2\) or \((250-04)^2\)= \(04^2=16\)
Last 2 digits of \(247^2\)or \((250-03)^2\)=\(03^2=09\)
Last 2 digits of \(248^2\)or \((250-02)^2= 02^2=04\)
Last 2 digits of \(249^2\) or \((250-01)^2= 01^2=01\)

last two digits of \((201*202*203*204*246*247*248*249)^2\)= last 2 digits of [01*04*09*16*16*09*04*01]=76


[quote="Bunuel"]What is the last two digits of \((201*202*203*204*246*247*248*249)^2\) ?

A. 10
B. 16
C. 36
D. 56
E. 76
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What is the last two digits of (201*202*203*204*246*247*248*249)^2 ? 19 Sep 2019, 09:13
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nick1816
Last 2 digits of \(201^2\)= \(01^2=01\)
Last 2 digits of \(202^2\)= \(02^2=04\)
Last 2 digits of \(203^2\)=\(03^2=09\)
Last 2 digits of \(204^2\)= \(04^2=16\)
Last 2 digits of \(246^2\) or \((250-04)^2\)= \(04^2=16\)
Last 2 digits of \(247^2\)or \((250-03)^2\)=\(03^2=09\)
Last 2 digits of \(248^2\)or \((250-02)^2= 02^2=04\)
Last 2 digits of \(249^2\) or \((250-01)^2= 01^2=01\)

last two digits of \((201*202*203*204*246*247*248*249)^2\)= last 2 digits of [01*04*09*16*16*09*04*01]=76


Bunuel
What is the last two digits of \((201*202*203*204*246*247*248*249)^2\) ?

A. 10
B. 16
C. 36
D. 56
E. 76
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Could you please explain how is (250-03)^2 = 03^2

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What is the last two digits of (201*202*203*204*246*247*248*249)^2 ? 19 Sep 2019, 09:30
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\((250-3)^2=250^2-2*250*3+03^2\)
\(250^2\)=62500 has 2 trailing 0's
2*250*3=1500 has 2 trailing 0's.

Hence last 2 digits of \((250-3)^2\) must be same as that of \(03^2\)

sachin1114GMCKS
nick1816
Last 2 digits of \(201^2\)= \(01^2=01\)
Last 2 digits of \(202^2\)= \(02^2=04\)
Last 2 digits of \(203^2\)=\(03^2=09\)
Last 2 digits of \(204^2\)= \(04^2=16\)
Last 2 digits of \(246^2\) or \((250-04)^2\)= \(04^2=16\)
Last 2 digits of \(247^2\)or \((250-03)^2\)=\(03^2=09\)
Last 2 digits of \(248^2\)or \((250-02)^2= 02^2=04\)
Last 2 digits of \(249^2\) or \((250-01)^2= 01^2=01\)

last two digits of \((201*202*203*204*246*247*248*249)^2\)= last 2 digits of [01*04*09*16*16*09*04*01]=76


Bunuel
What is the last two digits of \((201*202*203*204*246*247*248*249)^2\) ?

A. 10
B. 16
C. 36
D. 56
E. 76

Could you please explain how is (250-03)^2 = 03^2

Posted from my mobile device
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What is the last two digits of (201*202*203*204*246*247*248*249)^2 ? 02 Jan 2026, 01:46
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nick1816 for the last part of the question do we actually multiply all the numbers to arrive at 76 - 16*16*16*81? or is there any trick to that as well?
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What is the last two digits of (201*202*203*204*246*247*248*249)^2 ? 02 Jan 2026, 05:24
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1. Simplify the Numbers
First, we isolate the last two digits of each number in the product, as the hundreds digit (the "2" in 201 or 246) does not affect the last two digits of the result. The product becomes:
(01⋅02⋅03⋅04⋅46⋅47⋅48⋅49)^2


2. Multiply in Small Groups
We can calculate the last two digits by multiplying these in pairs and keeping only the last two digits at each step:
Group 1: 1⋅2⋅3⋅4=24
Group 2: 46⋅47=2162→ Last two digits are 62
Group 3: 48⋅49=2352→ Last two digits are 52


3. Combine the Groups
Now, we multiply these intermediate results together:
Step A: 24⋅62 = 1488→ Last two digits are 88
Step B: Multiply that result by the remaining group (88⋅52):
88⋅50 = 4400 (ends in 00)
88⋅2 = 176 (ends in 76)

4400+176 = 4576→ Last two digits are 76
The value inside the parentheses ends in 76.


4. Square the Final Result
Finally, we square the result to account for the exponent outside the parentheses:
76⋅76 = 5776
The last two digits are 76.

Why does 76 appear again? In mathematics, 76 is a special number called an automorphic number. When you multiply any number ending in 76 by another number ending in 76, the result will always end in 76.

Correct Answer: E. 76
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