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irakrogov76
@irakrogov76
August 2022
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Имеется семь последовательных натуральных чисел. Сумма первых трех равна 99. Чему равна сумма последних трех?
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N, n+1, n+2, n+3, n+4, n+5, n+6
n+n+1+n+2=99
3n+3=99
3(n+1)=99
n+1=99:3
n+1=33
n=33-1
n=32
n+4+n+5+n+6=3n+15
3n+15=3*32+15=96+15=111
Ответ: 111
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Verified answer
N, n+1, n+2, n+3, n+4, n+5, n+6n+n+1+n+2=99
3n+3=99
3(n+1)=99
n+1=99:3
n+1=33
n=33-1
n=32
n+4+n+5+n+6=3n+15
3n+15=3*32+15=96+15=111
Ответ: 111