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19 Sep 2019, 03:40
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Difficulty:
95%
(hard)
Question Stats:
38% (02:10) correct
62%
(01:58)
wrong
based on 55
sessions
History
Date
Time
Result
Not Attempted Yet
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
A. 10
B. 16
C. 36
D. 56
E. 76
Find the last 2 digits
(201*202*203*204*246*247*248*249)^2
1.36
2.56
3.76
4.16
(201*202*203*204*246*247*248*249)^2
1.36
2.56
3.76
4.16
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
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
19 Sep 2019, 09:13
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Bookmarks
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
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
A. 10
B. 16
C. 36
D. 56
E. 76
Could you please explain how is (250-03)^2 = 03^2
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